/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 225 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 64 ms]
(4) QTRS
(5) QTRSRRRProof [EQUIVALENT, 88 ms]
(6) QTRS
(7) DependencyPairsProof [EQUIVALENT, 115 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) AND
    (11) QDP
        (12) UsableRulesProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) QReductionProof [EQUIVALENT, 0 ms]
        (15) QDP
        (16) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (17) YES
    (18) QDP
        (19) UsableRulesProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QReductionProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (24) YES
    (25) QDP
        (26) UsableRulesProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) QReductionProof [EQUIVALENT, 0 ms]
        (29) QDP
        (30) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (31) YES
    (32) QDP
        (33) UsableRulesProof [EQUIVALENT, 0 ms]
        (34) QDP
        (35) QReductionProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (38) YES
    (39) QDP
        (40) UsableRulesProof [EQUIVALENT, 0 ms]
        (41) QDP
        (42) QReductionProof [EQUIVALENT, 0 ms]
        (43) QDP
        (44) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (45) YES
    (46) QDP
        (47) UsableRulesProof [EQUIVALENT, 0 ms]
        (48) QDP
        (49) QReductionProof [EQUIVALENT, 0 ms]
        (50) QDP
        (51) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (52) YES
    (53) QDP
        (54) UsableRulesProof [EQUIVALENT, 0 ms]
        (55) QDP
        (56) QReductionProof [EQUIVALENT, 0 ms]
        (57) QDP
        (58) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (59) YES
    (60) QDP
        (61) UsableRulesProof [EQUIVALENT, 0 ms]
        (62) QDP
        (63) QReductionProof [EQUIVALENT, 0 ms]
        (64) QDP
        (65) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (66) YES
    (67) QDP
        (68) MRRProof [EQUIVALENT, 86 ms]
        (69) QDP
        (70) MRRProof [EQUIVALENT, 108 ms]
        (71) QDP
        (72) MRRProof [EQUIVALENT, 125 ms]
        (73) QDP
        (74) MRRProof [EQUIVALENT, 112 ms]
        (75) QDP
        (76) QDPQMonotonicMRRProof [EQUIVALENT, 194 ms]
        (77) QDP
        (78) QDPOrderProof [EQUIVALENT, 90 ms]
        (79) QDP
        (80) DependencyGraphProof [EQUIVALENT, 0 ms]
        (81) QDP
        (82) QDPOrderProof [EQUIVALENT, 19 ms]
        (83) QDP
        (84) QDPOrderProof [EQUIVALENT, 8 ms]
        (85) QDP
        (86) QDPOrderProof [EQUIVALENT, 257 ms]
        (87) QDP
        (88) DependencyGraphProof [EQUIVALENT, 0 ms]
        (89) QDP
        (90) UsableRulesProof [EQUIVALENT, 0 ms]
        (91) QDP
        (92) QReductionProof [EQUIVALENT, 0 ms]
        (93) QDP
        (94) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (95) YES


----------------------------------------

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(0, XS)) -> mark(nil)
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(nil, XS)) -> mark(nil)
   active(zip(X, nil)) -> mark(nil)
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(tail(cons(X, XS))) -> mark(XS)
   active(repItems(nil)) -> mark(nil)
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))


----------------------------------------

(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(active(x_1)) = x_1
   POL(cons(x_1, x_2)) = x_1 + x_2
   POL(incr(x_1)) = 2*x_1
   POL(mark(x_1)) = x_1
   POL(nil) = 0
   POL(oddNs) = 0
   POL(pair(x_1, x_2)) = x_1 + x_2
   POL(pairNs) = 0
   POL(repItems(x_1)) = 2*x_1
   POL(s(x_1)) = x_1
   POL(tail(x_1)) = 2 + 2*x_1
   POL(take(x_1, x_2)) = x_1 + x_2
   POL(zip(x_1, x_2)) = 2*x_1 + 2*x_2
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   active(tail(cons(X, XS))) -> mark(XS)




----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(0, XS)) -> mark(nil)
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(nil, XS)) -> mark(nil)
   active(zip(X, nil)) -> mark(nil)
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(nil)) -> mark(nil)
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))


----------------------------------------

(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(active(x_1)) = x_1
   POL(cons(x_1, x_2)) = x_1 + x_2
   POL(incr(x_1)) = 2*x_1
   POL(mark(x_1)) = x_1
   POL(nil) = 0
   POL(oddNs) = 0
   POL(pair(x_1, x_2)) = x_1 + x_2
   POL(pairNs) = 0
   POL(repItems(x_1)) = 2*x_1
   POL(s(x_1)) = 2*x_1
   POL(tail(x_1)) = x_1
   POL(take(x_1, x_2)) = 1 + 2*x_1 + 2*x_2
   POL(zip(x_1, x_2)) = 2*x_1 + x_2
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   active(take(0, XS)) -> mark(nil)




----------------------------------------

(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(nil, XS)) -> mark(nil)
   active(zip(X, nil)) -> mark(nil)
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(nil)) -> mark(nil)
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))


----------------------------------------

(5) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(active(x_1)) = x_1
   POL(cons(x_1, x_2)) = 2*x_1 + x_2
   POL(incr(x_1)) = 2*x_1
   POL(mark(x_1)) = x_1
   POL(nil) = 2
   POL(oddNs) = 0
   POL(pair(x_1, x_2)) = x_1 + x_2
   POL(pairNs) = 0
   POL(repItems(x_1)) = 2*x_1
   POL(s(x_1)) = x_1
   POL(tail(x_1)) = x_1
   POL(take(x_1, x_2)) = 2*x_1 + x_2
   POL(zip(x_1, x_2)) = 2*x_1 + 2*x_2
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   active(zip(nil, XS)) -> mark(nil)
   active(zip(X, nil)) -> mark(nil)
   active(repItems(nil)) -> mark(nil)




----------------------------------------

(6)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))


----------------------------------------

(7) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVE(pairNs) -> MARK(cons(0, incr(oddNs)))
   ACTIVE(pairNs) -> CONS(0, incr(oddNs))
   ACTIVE(pairNs) -> INCR(oddNs)
   ACTIVE(oddNs) -> MARK(incr(pairNs))
   ACTIVE(oddNs) -> INCR(pairNs)
   ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS)))
   ACTIVE(incr(cons(X, XS))) -> CONS(s(X), incr(XS))
   ACTIVE(incr(cons(X, XS))) -> S(X)
   ACTIVE(incr(cons(X, XS))) -> INCR(XS)
   ACTIVE(take(s(N), cons(X, XS))) -> MARK(cons(X, take(N, XS)))
   ACTIVE(take(s(N), cons(X, XS))) -> CONS(X, take(N, XS))
   ACTIVE(take(s(N), cons(X, XS))) -> TAKE(N, XS)
   ACTIVE(zip(cons(X, XS), cons(Y, YS))) -> MARK(cons(pair(X, Y), zip(XS, YS)))
   ACTIVE(zip(cons(X, XS), cons(Y, YS))) -> CONS(pair(X, Y), zip(XS, YS))
   ACTIVE(zip(cons(X, XS), cons(Y, YS))) -> PAIR(X, Y)
   ACTIVE(zip(cons(X, XS), cons(Y, YS))) -> ZIP(XS, YS)
   ACTIVE(repItems(cons(X, XS))) -> MARK(cons(X, cons(X, repItems(XS))))
   ACTIVE(repItems(cons(X, XS))) -> CONS(X, cons(X, repItems(XS)))
   ACTIVE(repItems(cons(X, XS))) -> CONS(X, repItems(XS))
   ACTIVE(repItems(cons(X, XS))) -> REPITEMS(XS)
   MARK(pairNs) -> ACTIVE(pairNs)
   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
   MARK(cons(X1, X2)) -> CONS(mark(X1), X2)
   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(0) -> ACTIVE(0)
   MARK(incr(X)) -> ACTIVE(incr(mark(X)))
   MARK(incr(X)) -> INCR(mark(X))
   MARK(incr(X)) -> MARK(X)
   MARK(oddNs) -> ACTIVE(oddNs)
   MARK(s(X)) -> ACTIVE(s(mark(X)))
   MARK(s(X)) -> S(mark(X))
   MARK(s(X)) -> MARK(X)
   MARK(take(X1, X2)) -> ACTIVE(take(mark(X1), mark(X2)))
   MARK(take(X1, X2)) -> TAKE(mark(X1), mark(X2))
   MARK(take(X1, X2)) -> MARK(X1)
   MARK(take(X1, X2)) -> MARK(X2)
   MARK(nil) -> ACTIVE(nil)
   MARK(zip(X1, X2)) -> ACTIVE(zip(mark(X1), mark(X2)))
   MARK(zip(X1, X2)) -> ZIP(mark(X1), mark(X2))
   MARK(zip(X1, X2)) -> MARK(X1)
   MARK(zip(X1, X2)) -> MARK(X2)
   MARK(pair(X1, X2)) -> ACTIVE(pair(mark(X1), mark(X2)))
   MARK(pair(X1, X2)) -> PAIR(mark(X1), mark(X2))
   MARK(pair(X1, X2)) -> MARK(X1)
   MARK(pair(X1, X2)) -> MARK(X2)
   MARK(tail(X)) -> ACTIVE(tail(mark(X)))
   MARK(tail(X)) -> TAIL(mark(X))
   MARK(tail(X)) -> MARK(X)
   MARK(repItems(X)) -> ACTIVE(repItems(mark(X)))
   MARK(repItems(X)) -> REPITEMS(mark(X))
   MARK(repItems(X)) -> MARK(X)
   CONS(mark(X1), X2) -> CONS(X1, X2)
   CONS(X1, mark(X2)) -> CONS(X1, X2)
   CONS(active(X1), X2) -> CONS(X1, X2)
   CONS(X1, active(X2)) -> CONS(X1, X2)
   INCR(mark(X)) -> INCR(X)
   INCR(active(X)) -> INCR(X)
   S(mark(X)) -> S(X)
   S(active(X)) -> S(X)
   TAKE(mark(X1), X2) -> TAKE(X1, X2)
   TAKE(X1, mark(X2)) -> TAKE(X1, X2)
   TAKE(active(X1), X2) -> TAKE(X1, X2)
   TAKE(X1, active(X2)) -> TAKE(X1, X2)
   ZIP(mark(X1), X2) -> ZIP(X1, X2)
   ZIP(X1, mark(X2)) -> ZIP(X1, X2)
   ZIP(active(X1), X2) -> ZIP(X1, X2)
   ZIP(X1, active(X2)) -> ZIP(X1, X2)
   PAIR(mark(X1), X2) -> PAIR(X1, X2)
   PAIR(X1, mark(X2)) -> PAIR(X1, X2)
   PAIR(active(X1), X2) -> PAIR(X1, X2)
   PAIR(X1, active(X2)) -> PAIR(X1, X2)
   TAIL(mark(X)) -> TAIL(X)
   TAIL(active(X)) -> TAIL(X)
   REPITEMS(mark(X)) -> REPITEMS(X)
   REPITEMS(active(X)) -> REPITEMS(X)

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(9) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 9 SCCs with 24 less nodes.
----------------------------------------

(10)
Complex Obligation (AND)

----------------------------------------

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   REPITEMS(active(X)) -> REPITEMS(X)
   REPITEMS(mark(X)) -> REPITEMS(X)

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(12) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(13)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   REPITEMS(active(X)) -> REPITEMS(X)
   REPITEMS(mark(X)) -> REPITEMS(X)

R is empty.
The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(14) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))


----------------------------------------

(15)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   REPITEMS(active(X)) -> REPITEMS(X)
   REPITEMS(mark(X)) -> REPITEMS(X)

R is empty.
The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(16) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*REPITEMS(active(X)) -> REPITEMS(X)
The graph contains the following edges 1 > 1


*REPITEMS(mark(X)) -> REPITEMS(X)
The graph contains the following edges 1 > 1


----------------------------------------

(17)
YES

----------------------------------------

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TAIL(active(X)) -> TAIL(X)
   TAIL(mark(X)) -> TAIL(X)

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(19) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(20)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TAIL(active(X)) -> TAIL(X)
   TAIL(mark(X)) -> TAIL(X)

R is empty.
The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(21) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))


----------------------------------------

(22)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TAIL(active(X)) -> TAIL(X)
   TAIL(mark(X)) -> TAIL(X)

R is empty.
The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(23) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*TAIL(active(X)) -> TAIL(X)
The graph contains the following edges 1 > 1


*TAIL(mark(X)) -> TAIL(X)
The graph contains the following edges 1 > 1


----------------------------------------

(24)
YES

----------------------------------------

(25)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   PAIR(X1, mark(X2)) -> PAIR(X1, X2)
   PAIR(mark(X1), X2) -> PAIR(X1, X2)
   PAIR(active(X1), X2) -> PAIR(X1, X2)
   PAIR(X1, active(X2)) -> PAIR(X1, X2)

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(26) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(27)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   PAIR(X1, mark(X2)) -> PAIR(X1, X2)
   PAIR(mark(X1), X2) -> PAIR(X1, X2)
   PAIR(active(X1), X2) -> PAIR(X1, X2)
   PAIR(X1, active(X2)) -> PAIR(X1, X2)

R is empty.
The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(28) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))


----------------------------------------

(29)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   PAIR(X1, mark(X2)) -> PAIR(X1, X2)
   PAIR(mark(X1), X2) -> PAIR(X1, X2)
   PAIR(active(X1), X2) -> PAIR(X1, X2)
   PAIR(X1, active(X2)) -> PAIR(X1, X2)

R is empty.
The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(30) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*PAIR(X1, mark(X2)) -> PAIR(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2


*PAIR(mark(X1), X2) -> PAIR(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2


*PAIR(active(X1), X2) -> PAIR(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2


*PAIR(X1, active(X2)) -> PAIR(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2


----------------------------------------

(31)
YES

----------------------------------------

(32)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ZIP(X1, mark(X2)) -> ZIP(X1, X2)
   ZIP(mark(X1), X2) -> ZIP(X1, X2)
   ZIP(active(X1), X2) -> ZIP(X1, X2)
   ZIP(X1, active(X2)) -> ZIP(X1, X2)

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(33) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(34)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ZIP(X1, mark(X2)) -> ZIP(X1, X2)
   ZIP(mark(X1), X2) -> ZIP(X1, X2)
   ZIP(active(X1), X2) -> ZIP(X1, X2)
   ZIP(X1, active(X2)) -> ZIP(X1, X2)

R is empty.
The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(35) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))


----------------------------------------

(36)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ZIP(X1, mark(X2)) -> ZIP(X1, X2)
   ZIP(mark(X1), X2) -> ZIP(X1, X2)
   ZIP(active(X1), X2) -> ZIP(X1, X2)
   ZIP(X1, active(X2)) -> ZIP(X1, X2)

R is empty.
The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(37) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*ZIP(X1, mark(X2)) -> ZIP(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2


*ZIP(mark(X1), X2) -> ZIP(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2


*ZIP(active(X1), X2) -> ZIP(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2


*ZIP(X1, active(X2)) -> ZIP(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2


----------------------------------------

(38)
YES

----------------------------------------

(39)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TAKE(X1, mark(X2)) -> TAKE(X1, X2)
   TAKE(mark(X1), X2) -> TAKE(X1, X2)
   TAKE(active(X1), X2) -> TAKE(X1, X2)
   TAKE(X1, active(X2)) -> TAKE(X1, X2)

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(40) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(41)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TAKE(X1, mark(X2)) -> TAKE(X1, X2)
   TAKE(mark(X1), X2) -> TAKE(X1, X2)
   TAKE(active(X1), X2) -> TAKE(X1, X2)
   TAKE(X1, active(X2)) -> TAKE(X1, X2)

R is empty.
The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(42) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))


----------------------------------------

(43)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TAKE(X1, mark(X2)) -> TAKE(X1, X2)
   TAKE(mark(X1), X2) -> TAKE(X1, X2)
   TAKE(active(X1), X2) -> TAKE(X1, X2)
   TAKE(X1, active(X2)) -> TAKE(X1, X2)

R is empty.
The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(44) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*TAKE(X1, mark(X2)) -> TAKE(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2


*TAKE(mark(X1), X2) -> TAKE(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2


*TAKE(active(X1), X2) -> TAKE(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2


*TAKE(X1, active(X2)) -> TAKE(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2


----------------------------------------

(45)
YES

----------------------------------------

(46)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(active(X)) -> S(X)
   S(mark(X)) -> S(X)

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(47) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(48)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(active(X)) -> S(X)
   S(mark(X)) -> S(X)

R is empty.
The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(49) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))


----------------------------------------

(50)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(active(X)) -> S(X)
   S(mark(X)) -> S(X)

R is empty.
The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(51) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*S(active(X)) -> S(X)
The graph contains the following edges 1 > 1


*S(mark(X)) -> S(X)
The graph contains the following edges 1 > 1


----------------------------------------

(52)
YES

----------------------------------------

(53)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INCR(active(X)) -> INCR(X)
   INCR(mark(X)) -> INCR(X)

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(54) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(55)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INCR(active(X)) -> INCR(X)
   INCR(mark(X)) -> INCR(X)

R is empty.
The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(56) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))


----------------------------------------

(57)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INCR(active(X)) -> INCR(X)
   INCR(mark(X)) -> INCR(X)

R is empty.
The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(58) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*INCR(active(X)) -> INCR(X)
The graph contains the following edges 1 > 1


*INCR(mark(X)) -> INCR(X)
The graph contains the following edges 1 > 1


----------------------------------------

(59)
YES

----------------------------------------

(60)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   CONS(X1, mark(X2)) -> CONS(X1, X2)
   CONS(mark(X1), X2) -> CONS(X1, X2)
   CONS(active(X1), X2) -> CONS(X1, X2)
   CONS(X1, active(X2)) -> CONS(X1, X2)

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(61) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(62)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   CONS(X1, mark(X2)) -> CONS(X1, X2)
   CONS(mark(X1), X2) -> CONS(X1, X2)
   CONS(active(X1), X2) -> CONS(X1, X2)
   CONS(X1, active(X2)) -> CONS(X1, X2)

R is empty.
The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(63) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))


----------------------------------------

(64)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   CONS(X1, mark(X2)) -> CONS(X1, X2)
   CONS(mark(X1), X2) -> CONS(X1, X2)
   CONS(active(X1), X2) -> CONS(X1, X2)
   CONS(X1, active(X2)) -> CONS(X1, X2)

R is empty.
The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(65) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*CONS(X1, mark(X2)) -> CONS(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2


*CONS(mark(X1), X2) -> CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2


*CONS(active(X1), X2) -> CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2


*CONS(X1, active(X2)) -> CONS(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2


----------------------------------------

(66)
YES

----------------------------------------

(67)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
   ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS)))
   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(pairNs) -> ACTIVE(pairNs)
   ACTIVE(pairNs) -> MARK(cons(0, incr(oddNs)))
   MARK(incr(X)) -> ACTIVE(incr(mark(X)))
   ACTIVE(take(s(N), cons(X, XS))) -> MARK(cons(X, take(N, XS)))
   ACTIVE(zip(cons(X, XS), cons(Y, YS))) -> MARK(cons(pair(X, Y), zip(XS, YS)))
   ACTIVE(repItems(cons(X, XS))) -> MARK(cons(X, cons(X, repItems(XS))))
   MARK(incr(X)) -> MARK(X)
   MARK(oddNs) -> ACTIVE(oddNs)
   ACTIVE(oddNs) -> MARK(incr(pairNs))
   MARK(s(X)) -> ACTIVE(s(mark(X)))
   MARK(s(X)) -> MARK(X)
   MARK(take(X1, X2)) -> ACTIVE(take(mark(X1), mark(X2)))
   MARK(take(X1, X2)) -> MARK(X1)
   MARK(take(X1, X2)) -> MARK(X2)
   MARK(zip(X1, X2)) -> ACTIVE(zip(mark(X1), mark(X2)))
   MARK(zip(X1, X2)) -> MARK(X1)
   MARK(zip(X1, X2)) -> MARK(X2)
   MARK(pair(X1, X2)) -> ACTIVE(pair(mark(X1), mark(X2)))
   MARK(pair(X1, X2)) -> MARK(X1)
   MARK(pair(X1, X2)) -> MARK(X2)
   MARK(tail(X)) -> ACTIVE(tail(mark(X)))
   MARK(tail(X)) -> MARK(X)
   MARK(repItems(X)) -> ACTIVE(repItems(mark(X)))
   MARK(repItems(X)) -> MARK(X)

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(68) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   MARK(zip(X1, X2)) -> MARK(X1)
   MARK(zip(X1, X2)) -> MARK(X2)


Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(ACTIVE(x_1)) = 2*x_1
   POL(MARK(x_1)) = 2*x_1
   POL(active(x_1)) = x_1
   POL(cons(x_1, x_2)) = 2*x_1 + x_2
   POL(incr(x_1)) = 2*x_1
   POL(mark(x_1)) = x_1
   POL(nil) = 0
   POL(oddNs) = 0
   POL(pair(x_1, x_2)) = 2*x_1 + x_2
   POL(pairNs) = 0
   POL(repItems(x_1)) = 2*x_1
   POL(s(x_1)) = x_1
   POL(tail(x_1)) = 2*x_1
   POL(take(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(zip(x_1, x_2)) = 1 + 2*x_1 + 2*x_2


----------------------------------------

(69)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
   ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS)))
   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(pairNs) -> ACTIVE(pairNs)
   ACTIVE(pairNs) -> MARK(cons(0, incr(oddNs)))
   MARK(incr(X)) -> ACTIVE(incr(mark(X)))
   ACTIVE(take(s(N), cons(X, XS))) -> MARK(cons(X, take(N, XS)))
   ACTIVE(zip(cons(X, XS), cons(Y, YS))) -> MARK(cons(pair(X, Y), zip(XS, YS)))
   ACTIVE(repItems(cons(X, XS))) -> MARK(cons(X, cons(X, repItems(XS))))
   MARK(incr(X)) -> MARK(X)
   MARK(oddNs) -> ACTIVE(oddNs)
   ACTIVE(oddNs) -> MARK(incr(pairNs))
   MARK(s(X)) -> ACTIVE(s(mark(X)))
   MARK(s(X)) -> MARK(X)
   MARK(take(X1, X2)) -> ACTIVE(take(mark(X1), mark(X2)))
   MARK(take(X1, X2)) -> MARK(X1)
   MARK(take(X1, X2)) -> MARK(X2)
   MARK(zip(X1, X2)) -> ACTIVE(zip(mark(X1), mark(X2)))
   MARK(pair(X1, X2)) -> ACTIVE(pair(mark(X1), mark(X2)))
   MARK(pair(X1, X2)) -> MARK(X1)
   MARK(pair(X1, X2)) -> MARK(X2)
   MARK(tail(X)) -> ACTIVE(tail(mark(X)))
   MARK(tail(X)) -> MARK(X)
   MARK(repItems(X)) -> ACTIVE(repItems(mark(X)))
   MARK(repItems(X)) -> MARK(X)

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(70) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   MARK(take(X1, X2)) -> MARK(X1)
   MARK(take(X1, X2)) -> MARK(X2)


Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(ACTIVE(x_1)) = x_1
   POL(MARK(x_1)) = x_1
   POL(active(x_1)) = x_1
   POL(cons(x_1, x_2)) = 2*x_1 + x_2
   POL(incr(x_1)) = 2*x_1
   POL(mark(x_1)) = x_1
   POL(nil) = 0
   POL(oddNs) = 0
   POL(pair(x_1, x_2)) = 2*x_1 + x_2
   POL(pairNs) = 0
   POL(repItems(x_1)) = 2*x_1
   POL(s(x_1)) = 2*x_1
   POL(tail(x_1)) = x_1
   POL(take(x_1, x_2)) = 1 + 2*x_1 + x_2
   POL(zip(x_1, x_2)) = 2*x_1 + 2*x_2


----------------------------------------

(71)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
   ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS)))
   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(pairNs) -> ACTIVE(pairNs)
   ACTIVE(pairNs) -> MARK(cons(0, incr(oddNs)))
   MARK(incr(X)) -> ACTIVE(incr(mark(X)))
   ACTIVE(take(s(N), cons(X, XS))) -> MARK(cons(X, take(N, XS)))
   ACTIVE(zip(cons(X, XS), cons(Y, YS))) -> MARK(cons(pair(X, Y), zip(XS, YS)))
   ACTIVE(repItems(cons(X, XS))) -> MARK(cons(X, cons(X, repItems(XS))))
   MARK(incr(X)) -> MARK(X)
   MARK(oddNs) -> ACTIVE(oddNs)
   ACTIVE(oddNs) -> MARK(incr(pairNs))
   MARK(s(X)) -> ACTIVE(s(mark(X)))
   MARK(s(X)) -> MARK(X)
   MARK(take(X1, X2)) -> ACTIVE(take(mark(X1), mark(X2)))
   MARK(zip(X1, X2)) -> ACTIVE(zip(mark(X1), mark(X2)))
   MARK(pair(X1, X2)) -> ACTIVE(pair(mark(X1), mark(X2)))
   MARK(pair(X1, X2)) -> MARK(X1)
   MARK(pair(X1, X2)) -> MARK(X2)
   MARK(tail(X)) -> ACTIVE(tail(mark(X)))
   MARK(tail(X)) -> MARK(X)
   MARK(repItems(X)) -> ACTIVE(repItems(mark(X)))
   MARK(repItems(X)) -> MARK(X)

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(72) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   MARK(tail(X)) -> MARK(X)


Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(ACTIVE(x_1)) = 2*x_1
   POL(MARK(x_1)) = 2*x_1
   POL(active(x_1)) = x_1
   POL(cons(x_1, x_2)) = x_1 + x_2
   POL(incr(x_1)) = 2*x_1
   POL(mark(x_1)) = x_1
   POL(nil) = 0
   POL(oddNs) = 0
   POL(pair(x_1, x_2)) = x_1 + x_2
   POL(pairNs) = 0
   POL(repItems(x_1)) = 2*x_1
   POL(s(x_1)) = 2*x_1
   POL(tail(x_1)) = 1 + 2*x_1
   POL(take(x_1, x_2)) = x_1 + x_2
   POL(zip(x_1, x_2)) = 2*x_1 + 2*x_2


----------------------------------------

(73)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
   ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS)))
   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(pairNs) -> ACTIVE(pairNs)
   ACTIVE(pairNs) -> MARK(cons(0, incr(oddNs)))
   MARK(incr(X)) -> ACTIVE(incr(mark(X)))
   ACTIVE(take(s(N), cons(X, XS))) -> MARK(cons(X, take(N, XS)))
   ACTIVE(zip(cons(X, XS), cons(Y, YS))) -> MARK(cons(pair(X, Y), zip(XS, YS)))
   ACTIVE(repItems(cons(X, XS))) -> MARK(cons(X, cons(X, repItems(XS))))
   MARK(incr(X)) -> MARK(X)
   MARK(oddNs) -> ACTIVE(oddNs)
   ACTIVE(oddNs) -> MARK(incr(pairNs))
   MARK(s(X)) -> ACTIVE(s(mark(X)))
   MARK(s(X)) -> MARK(X)
   MARK(take(X1, X2)) -> ACTIVE(take(mark(X1), mark(X2)))
   MARK(zip(X1, X2)) -> ACTIVE(zip(mark(X1), mark(X2)))
   MARK(pair(X1, X2)) -> ACTIVE(pair(mark(X1), mark(X2)))
   MARK(pair(X1, X2)) -> MARK(X1)
   MARK(pair(X1, X2)) -> MARK(X2)
   MARK(tail(X)) -> ACTIVE(tail(mark(X)))
   MARK(repItems(X)) -> ACTIVE(repItems(mark(X)))
   MARK(repItems(X)) -> MARK(X)

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(74) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   MARK(repItems(X)) -> MARK(X)


Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(ACTIVE(x_1)) = 2*x_1
   POL(MARK(x_1)) = 2*x_1
   POL(active(x_1)) = x_1
   POL(cons(x_1, x_2)) = 2*x_1 + x_2
   POL(incr(x_1)) = 2*x_1
   POL(mark(x_1)) = x_1
   POL(nil) = 0
   POL(oddNs) = 0
   POL(pair(x_1, x_2)) = x_1 + x_2
   POL(pairNs) = 0
   POL(repItems(x_1)) = 2 + 2*x_1
   POL(s(x_1)) = 2*x_1
   POL(tail(x_1)) = x_1
   POL(take(x_1, x_2)) = x_1 + 2*x_2
   POL(zip(x_1, x_2)) = 2*x_1 + 2*x_2


----------------------------------------

(75)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
   ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS)))
   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(pairNs) -> ACTIVE(pairNs)
   ACTIVE(pairNs) -> MARK(cons(0, incr(oddNs)))
   MARK(incr(X)) -> ACTIVE(incr(mark(X)))
   ACTIVE(take(s(N), cons(X, XS))) -> MARK(cons(X, take(N, XS)))
   ACTIVE(zip(cons(X, XS), cons(Y, YS))) -> MARK(cons(pair(X, Y), zip(XS, YS)))
   ACTIVE(repItems(cons(X, XS))) -> MARK(cons(X, cons(X, repItems(XS))))
   MARK(incr(X)) -> MARK(X)
   MARK(oddNs) -> ACTIVE(oddNs)
   ACTIVE(oddNs) -> MARK(incr(pairNs))
   MARK(s(X)) -> ACTIVE(s(mark(X)))
   MARK(s(X)) -> MARK(X)
   MARK(take(X1, X2)) -> ACTIVE(take(mark(X1), mark(X2)))
   MARK(zip(X1, X2)) -> ACTIVE(zip(mark(X1), mark(X2)))
   MARK(pair(X1, X2)) -> ACTIVE(pair(mark(X1), mark(X2)))
   MARK(pair(X1, X2)) -> MARK(X1)
   MARK(pair(X1, X2)) -> MARK(X2)
   MARK(tail(X)) -> ACTIVE(tail(mark(X)))
   MARK(repItems(X)) -> ACTIVE(repItems(mark(X)))

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(76) QDPQMonotonicMRRProof (EQUIVALENT)
By using the Q-monotonic rule removal processor with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented such that it always occurs at a strongly monotonic position in a (P,Q,R)-chain.

Strictly oriented dependency pairs:

   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
   MARK(s(X)) -> ACTIVE(s(mark(X)))
   MARK(pair(X1, X2)) -> ACTIVE(pair(mark(X1), mark(X2)))
   MARK(tail(X)) -> ACTIVE(tail(mark(X)))


Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(ACTIVE(x_1)) = 2*x_1
   POL(MARK(x_1)) = 2
   POL(active(x_1)) = 0
   POL(cons(x_1, x_2)) = 0
   POL(incr(x_1)) = 1
   POL(mark(x_1)) = 0
   POL(nil) = 0
   POL(oddNs) = 1
   POL(pair(x_1, x_2)) = 0
   POL(pairNs) = 1
   POL(repItems(x_1)) = 1
   POL(s(x_1)) = 0
   POL(tail(x_1)) = 0
   POL(take(x_1, x_2)) = 1
   POL(zip(x_1, x_2)) = 1


----------------------------------------

(77)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS)))
   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(pairNs) -> ACTIVE(pairNs)
   ACTIVE(pairNs) -> MARK(cons(0, incr(oddNs)))
   MARK(incr(X)) -> ACTIVE(incr(mark(X)))
   ACTIVE(take(s(N), cons(X, XS))) -> MARK(cons(X, take(N, XS)))
   ACTIVE(zip(cons(X, XS), cons(Y, YS))) -> MARK(cons(pair(X, Y), zip(XS, YS)))
   ACTIVE(repItems(cons(X, XS))) -> MARK(cons(X, cons(X, repItems(XS))))
   MARK(incr(X)) -> MARK(X)
   MARK(oddNs) -> ACTIVE(oddNs)
   ACTIVE(oddNs) -> MARK(incr(pairNs))
   MARK(s(X)) -> MARK(X)
   MARK(take(X1, X2)) -> ACTIVE(take(mark(X1), mark(X2)))
   MARK(zip(X1, X2)) -> ACTIVE(zip(mark(X1), mark(X2)))
   MARK(pair(X1, X2)) -> MARK(X1)
   MARK(pair(X1, X2)) -> MARK(X2)
   MARK(repItems(X)) -> ACTIVE(repItems(mark(X)))

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(78) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   ACTIVE(pairNs) -> MARK(cons(0, incr(oddNs)))
   ACTIVE(zip(cons(X, XS), cons(Y, YS))) -> MARK(cons(pair(X, Y), zip(XS, YS)))
   ACTIVE(oddNs) -> MARK(incr(pairNs))
   MARK(pair(X1, X2)) -> MARK(X1)
   MARK(pair(X1, X2)) -> MARK(X2)
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
ACTIVE(x1)  =  x1

incr(x1)  =  x1

cons(x1, x2)  =  x1

MARK(x1)  =  x1

s(x1)  =  x1

pairNs  =  pairNs

0  =  0

mark(x1)  =  x1

take(x1, x2)  =  x2

zip(x1, x2)  =  zip(x1, x2)

pair(x1, x2)  =  pair(x1, x2)

repItems(x1)  =  x1

oddNs  =  oddNs

active(x1)  =  x1

nil  =  nil

tail(x1)  =  tail


Knuth-Bendix order [KBO] with precedence:oddNs > pairNs > 0

and weight map:

   oddNs=1
   tail=2
   zip_2=2
   0=1
   pair_2=1
   pairNs=1
   nil=2

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   mark(incr(X)) -> active(incr(mark(X)))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(0) -> active(0)
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   incr(active(X)) -> incr(X)
   incr(mark(X)) -> incr(X)
   take(X1, mark(X2)) -> take(X1, X2)
   take(mark(X1), X2) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   repItems(active(X)) -> repItems(X)
   repItems(mark(X)) -> repItems(X)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   s(active(X)) -> s(X)
   s(mark(X)) -> s(X)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(active(X)) -> tail(X)
   tail(mark(X)) -> tail(X)


----------------------------------------

(79)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS)))
   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(pairNs) -> ACTIVE(pairNs)
   MARK(incr(X)) -> ACTIVE(incr(mark(X)))
   ACTIVE(take(s(N), cons(X, XS))) -> MARK(cons(X, take(N, XS)))
   ACTIVE(repItems(cons(X, XS))) -> MARK(cons(X, cons(X, repItems(XS))))
   MARK(incr(X)) -> MARK(X)
   MARK(oddNs) -> ACTIVE(oddNs)
   MARK(s(X)) -> MARK(X)
   MARK(take(X1, X2)) -> ACTIVE(take(mark(X1), mark(X2)))
   MARK(zip(X1, X2)) -> ACTIVE(zip(mark(X1), mark(X2)))
   MARK(repItems(X)) -> ACTIVE(repItems(mark(X)))

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(80) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

(81)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(incr(X)) -> ACTIVE(incr(mark(X)))
   ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS)))
   ACTIVE(take(s(N), cons(X, XS))) -> MARK(cons(X, take(N, XS)))
   ACTIVE(repItems(cons(X, XS))) -> MARK(cons(X, cons(X, repItems(XS))))
   MARK(incr(X)) -> MARK(X)
   MARK(s(X)) -> MARK(X)
   MARK(take(X1, X2)) -> ACTIVE(take(mark(X1), mark(X2)))
   MARK(zip(X1, X2)) -> ACTIVE(zip(mark(X1), mark(X2)))
   MARK(repItems(X)) -> ACTIVE(repItems(mark(X)))

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(82) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   ACTIVE(repItems(cons(X, XS))) -> MARK(cons(X, cons(X, repItems(XS))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
MARK(x1)  =  x1

cons(x1, x2)  =  x1

incr(x1)  =  x1

ACTIVE(x1)  =  x1

mark(x1)  =  x1

s(x1)  =  x1

take(x1, x2)  =  x2

repItems(x1)  =  repItems(x1)

zip(x1, x2)  =  zip

pairNs  =  pairNs

active(x1)  =  x1

0  =  0

oddNs  =  oddNs

pair(x1, x2)  =  pair

nil  =  nil

tail(x1)  =  tail


Knuth-Bendix order [KBO] with precedence:oddNs > pairNs > 0

and weight map:

   oddNs=1
   tail=2
   zip=3
   0=1
   pair=2
   pairNs=1
   repItems_1=1
   nil=2

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   mark(incr(X)) -> active(incr(mark(X)))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(0) -> active(0)
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   incr(active(X)) -> incr(X)
   incr(mark(X)) -> incr(X)
   take(X1, mark(X2)) -> take(X1, X2)
   take(mark(X1), X2) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   repItems(active(X)) -> repItems(X)
   repItems(mark(X)) -> repItems(X)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   s(active(X)) -> s(X)
   s(mark(X)) -> s(X)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(active(X)) -> tail(X)
   tail(mark(X)) -> tail(X)


----------------------------------------

(83)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(incr(X)) -> ACTIVE(incr(mark(X)))
   ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS)))
   ACTIVE(take(s(N), cons(X, XS))) -> MARK(cons(X, take(N, XS)))
   MARK(incr(X)) -> MARK(X)
   MARK(s(X)) -> MARK(X)
   MARK(take(X1, X2)) -> ACTIVE(take(mark(X1), mark(X2)))
   MARK(zip(X1, X2)) -> ACTIVE(zip(mark(X1), mark(X2)))
   MARK(repItems(X)) -> ACTIVE(repItems(mark(X)))

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(84) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   ACTIVE(take(s(N), cons(X, XS))) -> MARK(cons(X, take(N, XS)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
MARK(x1)  =  x1

cons(x1, x2)  =  x1

incr(x1)  =  x1

ACTIVE(x1)  =  x1

mark(x1)  =  x1

s(x1)  =  x1

take(x1, x2)  =  take(x2)

zip(x1, x2)  =  zip

repItems(x1)  =  x1

pairNs  =  pairNs

active(x1)  =  x1

0  =  0

oddNs  =  oddNs

pair(x1, x2)  =  pair

nil  =  nil

tail(x1)  =  tail


Knuth-Bendix order [KBO] with precedence:oddNs > pairNs > 0

and weight map:

   oddNs=1
   tail=2
   zip=3
   take_1=1
   0=1
   pair=2
   pairNs=1
   nil=2

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   mark(incr(X)) -> active(incr(mark(X)))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(0) -> active(0)
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   incr(active(X)) -> incr(X)
   incr(mark(X)) -> incr(X)
   take(X1, mark(X2)) -> take(X1, X2)
   take(mark(X1), X2) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   repItems(active(X)) -> repItems(X)
   repItems(mark(X)) -> repItems(X)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   s(active(X)) -> s(X)
   s(mark(X)) -> s(X)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(active(X)) -> tail(X)
   tail(mark(X)) -> tail(X)


----------------------------------------

(85)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(incr(X)) -> ACTIVE(incr(mark(X)))
   ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS)))
   MARK(incr(X)) -> MARK(X)
   MARK(s(X)) -> MARK(X)
   MARK(take(X1, X2)) -> ACTIVE(take(mark(X1), mark(X2)))
   MARK(zip(X1, X2)) -> ACTIVE(zip(mark(X1), mark(X2)))
   MARK(repItems(X)) -> ACTIVE(repItems(mark(X)))

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(86) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS)))
   MARK(incr(X)) -> MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
MARK(x1)  =  x1

cons(x1, x2)  =  x1

incr(x1)  =  incr(x1)

ACTIVE(x1)  =  x1

mark(x1)  =  x1

s(x1)  =  x1

take(x1, x2)  =  x2

zip(x1, x2)  =  zip

repItems(x1)  =  x1

pairNs  =  pairNs

active(x1)  =  x1

0  =  0

oddNs  =  oddNs

pair(x1, x2)  =  pair

nil  =  nil

tail(x1)  =  tail


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   oddNs=5
   tail=1
   zip=3
   0=2
   pair=2
   pairNs=3
   incr_1=1
   nil=2

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   mark(incr(X)) -> active(incr(mark(X)))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(0) -> active(0)
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   incr(active(X)) -> incr(X)
   incr(mark(X)) -> incr(X)
   take(X1, mark(X2)) -> take(X1, X2)
   take(mark(X1), X2) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   repItems(active(X)) -> repItems(X)
   repItems(mark(X)) -> repItems(X)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   s(active(X)) -> s(X)
   s(mark(X)) -> s(X)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(active(X)) -> tail(X)
   tail(mark(X)) -> tail(X)


----------------------------------------

(87)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(incr(X)) -> ACTIVE(incr(mark(X)))
   MARK(s(X)) -> MARK(X)
   MARK(take(X1, X2)) -> ACTIVE(take(mark(X1), mark(X2)))
   MARK(zip(X1, X2)) -> ACTIVE(zip(mark(X1), mark(X2)))
   MARK(repItems(X)) -> ACTIVE(repItems(mark(X)))

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(88) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.
----------------------------------------

(89)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(s(X)) -> MARK(X)
   MARK(cons(X1, X2)) -> MARK(X1)

The TRS R consists of the following rules:

   active(pairNs) -> mark(cons(0, incr(oddNs)))
   active(oddNs) -> mark(incr(pairNs))
   active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS)))
   active(take(s(N), cons(X, XS))) -> mark(cons(X, take(N, XS)))
   active(zip(cons(X, XS), cons(Y, YS))) -> mark(cons(pair(X, Y), zip(XS, YS)))
   active(repItems(cons(X, XS))) -> mark(cons(X, cons(X, repItems(XS))))
   mark(pairNs) -> active(pairNs)
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(0) -> active(0)
   mark(incr(X)) -> active(incr(mark(X)))
   mark(oddNs) -> active(oddNs)
   mark(s(X)) -> active(s(mark(X)))
   mark(take(X1, X2)) -> active(take(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   mark(zip(X1, X2)) -> active(zip(mark(X1), mark(X2)))
   mark(pair(X1, X2)) -> active(pair(mark(X1), mark(X2)))
   mark(tail(X)) -> active(tail(mark(X)))
   mark(repItems(X)) -> active(repItems(mark(X)))
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   incr(mark(X)) -> incr(X)
   incr(active(X)) -> incr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   take(mark(X1), X2) -> take(X1, X2)
   take(X1, mark(X2)) -> take(X1, X2)
   take(active(X1), X2) -> take(X1, X2)
   take(X1, active(X2)) -> take(X1, X2)
   zip(mark(X1), X2) -> zip(X1, X2)
   zip(X1, mark(X2)) -> zip(X1, X2)
   zip(active(X1), X2) -> zip(X1, X2)
   zip(X1, active(X2)) -> zip(X1, X2)
   pair(mark(X1), X2) -> pair(X1, X2)
   pair(X1, mark(X2)) -> pair(X1, X2)
   pair(active(X1), X2) -> pair(X1, X2)
   pair(X1, active(X2)) -> pair(X1, X2)
   tail(mark(X)) -> tail(X)
   tail(active(X)) -> tail(X)
   repItems(mark(X)) -> repItems(X)
   repItems(active(X)) -> repItems(X)

The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(90) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(91)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(s(X)) -> MARK(X)
   MARK(cons(X1, X2)) -> MARK(X1)

R is empty.
The set Q consists of the following terms:

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   incr(mark(x0))
   incr(active(x0))
   s(mark(x0))
   s(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(92) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   active(pairNs)
   active(oddNs)
   active(incr(cons(x0, x1)))
   active(take(0, x0))
   active(take(s(x0), cons(x1, x2)))
   active(zip(nil, x0))
   active(zip(x0, nil))
   active(zip(cons(x0, x1), cons(x2, x3)))
   active(tail(cons(x0, x1)))
   active(repItems(nil))
   active(repItems(cons(x0, x1)))
   mark(pairNs)
   mark(cons(x0, x1))
   mark(0)
   mark(incr(x0))
   mark(oddNs)
   mark(s(x0))
   mark(take(x0, x1))
   mark(nil)
   mark(zip(x0, x1))
   mark(pair(x0, x1))
   mark(tail(x0))
   mark(repItems(x0))
   incr(mark(x0))
   incr(active(x0))
   take(mark(x0), x1)
   take(x0, mark(x1))
   take(active(x0), x1)
   take(x0, active(x1))
   zip(mark(x0), x1)
   zip(x0, mark(x1))
   zip(active(x0), x1)
   zip(x0, active(x1))
   pair(mark(x0), x1)
   pair(x0, mark(x1))
   pair(active(x0), x1)
   pair(x0, active(x1))
   tail(mark(x0))
   tail(active(x0))
   repItems(mark(x0))
   repItems(active(x0))


----------------------------------------

(93)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(s(X)) -> MARK(X)
   MARK(cons(X1, X2)) -> MARK(X1)

R is empty.
The set Q consists of the following terms:

   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   s(mark(x0))
   s(active(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(94) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*MARK(s(X)) -> MARK(X)
The graph contains the following edges 1 > 1


*MARK(cons(X1, X2)) -> MARK(X1)
The graph contains the following edges 1 > 1


----------------------------------------

(95)
YES