/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 41 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QReductionProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) UsableRulesProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) QReductionProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (18) YES
    (19) QDP
        (20) UsableRulesProof [EQUIVALENT, 0 ms]
        (21) QDP
        (22) QReductionProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (25) YES
    (26) QDP
        (27) UsableRulesProof [EQUIVALENT, 0 ms]
        (28) QDP
        (29) QReductionProof [EQUIVALENT, 0 ms]
        (30) QDP
        (31) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (32) YES
    (33) QDP
        (34) UsableRulesProof [EQUIVALENT, 0 ms]
        (35) QDP
        (36) QReductionProof [EQUIVALENT, 0 ms]
        (37) QDP
        (38) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (39) YES
    (40) QDP
        (41) UsableRulesProof [EQUIVALENT, 0 ms]
        (42) QDP
        (43) QReductionProof [EQUIVALENT, 0 ms]
        (44) QDP
        (45) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (46) YES
    (47) QDP
        (48) UsableRulesProof [EQUIVALENT, 0 ms]
        (49) QDP
        (50) QReductionProof [EQUIVALENT, 0 ms]
        (51) QDP
        (52) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (53) YES
    (54) QDP
        (55) UsableRulesProof [EQUIVALENT, 0 ms]
        (56) QDP
        (57) QReductionProof [EQUIVALENT, 0 ms]
        (58) QDP
        (59) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (60) YES
    (61) QDP
        (62) QDPOrderProof [EQUIVALENT, 293 ms]
        (63) QDP
        (64) QDPOrderProof [EQUIVALENT, 275 ms]
        (65) QDP
        (66) UsableRulesProof [EQUIVALENT, 0 ms]
        (67) QDP
        (68) QReductionProof [EQUIVALENT, 0 ms]
        (69) QDP
        (70) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (71) YES


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

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

   active(terms(N)) -> mark(cons(recip(sqr(N)), terms(s(N))))
   active(sqr(0)) -> mark(0)
   active(sqr(s(X))) -> mark(s(add(sqr(X), dbl(X))))
   active(dbl(0)) -> mark(0)
   active(dbl(s(X))) -> mark(s(s(dbl(X))))
   active(add(0, X)) -> mark(X)
   active(add(s(X), Y)) -> mark(s(add(X, Y)))
   active(first(0, X)) -> mark(nil)
   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
   mark(terms(X)) -> active(terms(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(recip(X)) -> active(recip(mark(X)))
   mark(sqr(X)) -> active(sqr(mark(X)))
   mark(s(X)) -> active(s(mark(X)))
   mark(0) -> active(0)
   mark(add(X1, X2)) -> active(add(mark(X1), mark(X2)))
   mark(dbl(X)) -> active(dbl(mark(X)))
   mark(first(X1, X2)) -> active(first(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   terms(mark(X)) -> terms(X)
   terms(active(X)) -> terms(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)
   recip(mark(X)) -> recip(X)
   recip(active(X)) -> recip(X)
   sqr(mark(X)) -> sqr(X)
   sqr(active(X)) -> sqr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   add(mark(X1), X2) -> add(X1, X2)
   add(X1, mark(X2)) -> add(X1, X2)
   add(active(X1), X2) -> add(X1, X2)
   add(X1, active(X2)) -> add(X1, X2)
   dbl(mark(X)) -> dbl(X)
   dbl(active(X)) -> dbl(X)
   first(mark(X1), X2) -> first(X1, X2)
   first(X1, mark(X2)) -> first(X1, X2)
   first(active(X1), X2) -> first(X1, X2)
   first(X1, active(X2)) -> first(X1, X2)

The set Q consists of the following terms:

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))


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

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

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVE(terms(N)) -> MARK(cons(recip(sqr(N)), terms(s(N))))
   ACTIVE(terms(N)) -> CONS(recip(sqr(N)), terms(s(N)))
   ACTIVE(terms(N)) -> RECIP(sqr(N))
   ACTIVE(terms(N)) -> SQR(N)
   ACTIVE(terms(N)) -> TERMS(s(N))
   ACTIVE(terms(N)) -> S(N)
   ACTIVE(sqr(0)) -> MARK(0)
   ACTIVE(sqr(s(X))) -> MARK(s(add(sqr(X), dbl(X))))
   ACTIVE(sqr(s(X))) -> S(add(sqr(X), dbl(X)))
   ACTIVE(sqr(s(X))) -> ADD(sqr(X), dbl(X))
   ACTIVE(sqr(s(X))) -> SQR(X)
   ACTIVE(sqr(s(X))) -> DBL(X)
   ACTIVE(dbl(0)) -> MARK(0)
   ACTIVE(dbl(s(X))) -> MARK(s(s(dbl(X))))
   ACTIVE(dbl(s(X))) -> S(s(dbl(X)))
   ACTIVE(dbl(s(X))) -> S(dbl(X))
   ACTIVE(dbl(s(X))) -> DBL(X)
   ACTIVE(add(0, X)) -> MARK(X)
   ACTIVE(add(s(X), Y)) -> MARK(s(add(X, Y)))
   ACTIVE(add(s(X), Y)) -> S(add(X, Y))
   ACTIVE(add(s(X), Y)) -> ADD(X, Y)
   ACTIVE(first(0, X)) -> MARK(nil)
   ACTIVE(first(s(X), cons(Y, Z))) -> MARK(cons(Y, first(X, Z)))
   ACTIVE(first(s(X), cons(Y, Z))) -> CONS(Y, first(X, Z))
   ACTIVE(first(s(X), cons(Y, Z))) -> FIRST(X, Z)
   MARK(terms(X)) -> ACTIVE(terms(mark(X)))
   MARK(terms(X)) -> TERMS(mark(X))
   MARK(terms(X)) -> MARK(X)
   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
   MARK(cons(X1, X2)) -> CONS(mark(X1), X2)
   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(recip(X)) -> ACTIVE(recip(mark(X)))
   MARK(recip(X)) -> RECIP(mark(X))
   MARK(recip(X)) -> MARK(X)
   MARK(sqr(X)) -> ACTIVE(sqr(mark(X)))
   MARK(sqr(X)) -> SQR(mark(X))
   MARK(sqr(X)) -> MARK(X)
   MARK(s(X)) -> ACTIVE(s(mark(X)))
   MARK(s(X)) -> S(mark(X))
   MARK(s(X)) -> MARK(X)
   MARK(0) -> ACTIVE(0)
   MARK(add(X1, X2)) -> ACTIVE(add(mark(X1), mark(X2)))
   MARK(add(X1, X2)) -> ADD(mark(X1), mark(X2))
   MARK(add(X1, X2)) -> MARK(X1)
   MARK(add(X1, X2)) -> MARK(X2)
   MARK(dbl(X)) -> ACTIVE(dbl(mark(X)))
   MARK(dbl(X)) -> DBL(mark(X))
   MARK(dbl(X)) -> MARK(X)
   MARK(first(X1, X2)) -> ACTIVE(first(mark(X1), mark(X2)))
   MARK(first(X1, X2)) -> FIRST(mark(X1), mark(X2))
   MARK(first(X1, X2)) -> MARK(X1)
   MARK(first(X1, X2)) -> MARK(X2)
   MARK(nil) -> ACTIVE(nil)
   TERMS(mark(X)) -> TERMS(X)
   TERMS(active(X)) -> TERMS(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)
   RECIP(mark(X)) -> RECIP(X)
   RECIP(active(X)) -> RECIP(X)
   SQR(mark(X)) -> SQR(X)
   SQR(active(X)) -> SQR(X)
   S(mark(X)) -> S(X)
   S(active(X)) -> S(X)
   ADD(mark(X1), X2) -> ADD(X1, X2)
   ADD(X1, mark(X2)) -> ADD(X1, X2)
   ADD(active(X1), X2) -> ADD(X1, X2)
   ADD(X1, active(X2)) -> ADD(X1, X2)
   DBL(mark(X)) -> DBL(X)
   DBL(active(X)) -> DBL(X)
   FIRST(mark(X1), X2) -> FIRST(X1, X2)
   FIRST(X1, mark(X2)) -> FIRST(X1, X2)
   FIRST(active(X1), X2) -> FIRST(X1, X2)
   FIRST(X1, active(X2)) -> FIRST(X1, X2)

The TRS R consists of the following rules:

   active(terms(N)) -> mark(cons(recip(sqr(N)), terms(s(N))))
   active(sqr(0)) -> mark(0)
   active(sqr(s(X))) -> mark(s(add(sqr(X), dbl(X))))
   active(dbl(0)) -> mark(0)
   active(dbl(s(X))) -> mark(s(s(dbl(X))))
   active(add(0, X)) -> mark(X)
   active(add(s(X), Y)) -> mark(s(add(X, Y)))
   active(first(0, X)) -> mark(nil)
   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
   mark(terms(X)) -> active(terms(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(recip(X)) -> active(recip(mark(X)))
   mark(sqr(X)) -> active(sqr(mark(X)))
   mark(s(X)) -> active(s(mark(X)))
   mark(0) -> active(0)
   mark(add(X1, X2)) -> active(add(mark(X1), mark(X2)))
   mark(dbl(X)) -> active(dbl(mark(X)))
   mark(first(X1, X2)) -> active(first(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   terms(mark(X)) -> terms(X)
   terms(active(X)) -> terms(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)
   recip(mark(X)) -> recip(X)
   recip(active(X)) -> recip(X)
   sqr(mark(X)) -> sqr(X)
   sqr(active(X)) -> sqr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   add(mark(X1), X2) -> add(X1, X2)
   add(X1, mark(X2)) -> add(X1, X2)
   add(active(X1), X2) -> add(X1, X2)
   add(X1, active(X2)) -> add(X1, X2)
   dbl(mark(X)) -> dbl(X)
   dbl(active(X)) -> dbl(X)
   first(mark(X1), X2) -> first(X1, X2)
   first(X1, mark(X2)) -> first(X1, X2)
   first(active(X1), X2) -> first(X1, X2)
   first(X1, active(X2)) -> first(X1, X2)

The set Q consists of the following terms:

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 9 SCCs with 29 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   active(terms(N)) -> mark(cons(recip(sqr(N)), terms(s(N))))
   active(sqr(0)) -> mark(0)
   active(sqr(s(X))) -> mark(s(add(sqr(X), dbl(X))))
   active(dbl(0)) -> mark(0)
   active(dbl(s(X))) -> mark(s(s(dbl(X))))
   active(add(0, X)) -> mark(X)
   active(add(s(X), Y)) -> mark(s(add(X, Y)))
   active(first(0, X)) -> mark(nil)
   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
   mark(terms(X)) -> active(terms(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(recip(X)) -> active(recip(mark(X)))
   mark(sqr(X)) -> active(sqr(mark(X)))
   mark(s(X)) -> active(s(mark(X)))
   mark(0) -> active(0)
   mark(add(X1, X2)) -> active(add(mark(X1), mark(X2)))
   mark(dbl(X)) -> active(dbl(mark(X)))
   mark(first(X1, X2)) -> active(first(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   terms(mark(X)) -> terms(X)
   terms(active(X)) -> terms(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)
   recip(mark(X)) -> recip(X)
   recip(active(X)) -> recip(X)
   sqr(mark(X)) -> sqr(X)
   sqr(active(X)) -> sqr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   add(mark(X1), X2) -> add(X1, X2)
   add(X1, mark(X2)) -> add(X1, X2)
   add(active(X1), X2) -> add(X1, X2)
   add(X1, active(X2)) -> add(X1, X2)
   dbl(mark(X)) -> dbl(X)
   dbl(active(X)) -> dbl(X)
   first(mark(X1), X2) -> first(X1, X2)
   first(X1, mark(X2)) -> first(X1, X2)
   first(active(X1), X2) -> first(X1, X2)
   first(X1, active(X2)) -> first(X1, X2)

The set Q consists of the following terms:

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

(6) 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.
----------------------------------------

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

(8) 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].

   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))


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

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)

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

(10) 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:
*FIRST(X1, mark(X2)) -> FIRST(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2


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


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


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


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

(11)
YES

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

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DBL(active(X)) -> DBL(X)
   DBL(mark(X)) -> DBL(X)

The TRS R consists of the following rules:

   active(terms(N)) -> mark(cons(recip(sqr(N)), terms(s(N))))
   active(sqr(0)) -> mark(0)
   active(sqr(s(X))) -> mark(s(add(sqr(X), dbl(X))))
   active(dbl(0)) -> mark(0)
   active(dbl(s(X))) -> mark(s(s(dbl(X))))
   active(add(0, X)) -> mark(X)
   active(add(s(X), Y)) -> mark(s(add(X, Y)))
   active(first(0, X)) -> mark(nil)
   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
   mark(terms(X)) -> active(terms(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(recip(X)) -> active(recip(mark(X)))
   mark(sqr(X)) -> active(sqr(mark(X)))
   mark(s(X)) -> active(s(mark(X)))
   mark(0) -> active(0)
   mark(add(X1, X2)) -> active(add(mark(X1), mark(X2)))
   mark(dbl(X)) -> active(dbl(mark(X)))
   mark(first(X1, X2)) -> active(first(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   terms(mark(X)) -> terms(X)
   terms(active(X)) -> terms(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)
   recip(mark(X)) -> recip(X)
   recip(active(X)) -> recip(X)
   sqr(mark(X)) -> sqr(X)
   sqr(active(X)) -> sqr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   add(mark(X1), X2) -> add(X1, X2)
   add(X1, mark(X2)) -> add(X1, X2)
   add(active(X1), X2) -> add(X1, X2)
   add(X1, active(X2)) -> add(X1, X2)
   dbl(mark(X)) -> dbl(X)
   dbl(active(X)) -> dbl(X)
   first(mark(X1), X2) -> first(X1, X2)
   first(X1, mark(X2)) -> first(X1, X2)
   first(active(X1), X2) -> first(X1, X2)
   first(X1, active(X2)) -> first(X1, X2)

The set Q consists of the following terms:

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

(13) 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.
----------------------------------------

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DBL(active(X)) -> DBL(X)
   DBL(mark(X)) -> DBL(X)

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

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

(15) 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].

   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))


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

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DBL(active(X)) -> DBL(X)
   DBL(mark(X)) -> DBL(X)

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

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)

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

(17) 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:
*DBL(active(X)) -> DBL(X)
The graph contains the following edges 1 > 1


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


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

(18)
YES

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

(19)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   active(terms(N)) -> mark(cons(recip(sqr(N)), terms(s(N))))
   active(sqr(0)) -> mark(0)
   active(sqr(s(X))) -> mark(s(add(sqr(X), dbl(X))))
   active(dbl(0)) -> mark(0)
   active(dbl(s(X))) -> mark(s(s(dbl(X))))
   active(add(0, X)) -> mark(X)
   active(add(s(X), Y)) -> mark(s(add(X, Y)))
   active(first(0, X)) -> mark(nil)
   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
   mark(terms(X)) -> active(terms(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(recip(X)) -> active(recip(mark(X)))
   mark(sqr(X)) -> active(sqr(mark(X)))
   mark(s(X)) -> active(s(mark(X)))
   mark(0) -> active(0)
   mark(add(X1, X2)) -> active(add(mark(X1), mark(X2)))
   mark(dbl(X)) -> active(dbl(mark(X)))
   mark(first(X1, X2)) -> active(first(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   terms(mark(X)) -> terms(X)
   terms(active(X)) -> terms(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)
   recip(mark(X)) -> recip(X)
   recip(active(X)) -> recip(X)
   sqr(mark(X)) -> sqr(X)
   sqr(active(X)) -> sqr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   add(mark(X1), X2) -> add(X1, X2)
   add(X1, mark(X2)) -> add(X1, X2)
   add(active(X1), X2) -> add(X1, X2)
   add(X1, active(X2)) -> add(X1, X2)
   dbl(mark(X)) -> dbl(X)
   dbl(active(X)) -> dbl(X)
   first(mark(X1), X2) -> first(X1, X2)
   first(X1, mark(X2)) -> first(X1, X2)
   first(active(X1), X2) -> first(X1, X2)
   first(X1, active(X2)) -> first(X1, X2)

The set Q consists of the following terms:

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

(20) 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.
----------------------------------------

(21)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

(22) 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].

   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))


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

(23)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)

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

(24) 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:
*ADD(X1, mark(X2)) -> ADD(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2


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


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


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


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

(25)
YES

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

(26)
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(terms(N)) -> mark(cons(recip(sqr(N)), terms(s(N))))
   active(sqr(0)) -> mark(0)
   active(sqr(s(X))) -> mark(s(add(sqr(X), dbl(X))))
   active(dbl(0)) -> mark(0)
   active(dbl(s(X))) -> mark(s(s(dbl(X))))
   active(add(0, X)) -> mark(X)
   active(add(s(X), Y)) -> mark(s(add(X, Y)))
   active(first(0, X)) -> mark(nil)
   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
   mark(terms(X)) -> active(terms(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(recip(X)) -> active(recip(mark(X)))
   mark(sqr(X)) -> active(sqr(mark(X)))
   mark(s(X)) -> active(s(mark(X)))
   mark(0) -> active(0)
   mark(add(X1, X2)) -> active(add(mark(X1), mark(X2)))
   mark(dbl(X)) -> active(dbl(mark(X)))
   mark(first(X1, X2)) -> active(first(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   terms(mark(X)) -> terms(X)
   terms(active(X)) -> terms(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)
   recip(mark(X)) -> recip(X)
   recip(active(X)) -> recip(X)
   sqr(mark(X)) -> sqr(X)
   sqr(active(X)) -> sqr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   add(mark(X1), X2) -> add(X1, X2)
   add(X1, mark(X2)) -> add(X1, X2)
   add(active(X1), X2) -> add(X1, X2)
   add(X1, active(X2)) -> add(X1, X2)
   dbl(mark(X)) -> dbl(X)
   dbl(active(X)) -> dbl(X)
   first(mark(X1), X2) -> first(X1, X2)
   first(X1, mark(X2)) -> first(X1, X2)
   first(active(X1), X2) -> first(X1, X2)
   first(X1, active(X2)) -> first(X1, X2)

The set Q consists of the following terms:

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

(27) 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.
----------------------------------------

(28)
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(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

(29) 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].

   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))


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

(30)
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(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)

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

(31) 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


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

(32)
YES

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

(33)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SQR(active(X)) -> SQR(X)
   SQR(mark(X)) -> SQR(X)

The TRS R consists of the following rules:

   active(terms(N)) -> mark(cons(recip(sqr(N)), terms(s(N))))
   active(sqr(0)) -> mark(0)
   active(sqr(s(X))) -> mark(s(add(sqr(X), dbl(X))))
   active(dbl(0)) -> mark(0)
   active(dbl(s(X))) -> mark(s(s(dbl(X))))
   active(add(0, X)) -> mark(X)
   active(add(s(X), Y)) -> mark(s(add(X, Y)))
   active(first(0, X)) -> mark(nil)
   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
   mark(terms(X)) -> active(terms(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(recip(X)) -> active(recip(mark(X)))
   mark(sqr(X)) -> active(sqr(mark(X)))
   mark(s(X)) -> active(s(mark(X)))
   mark(0) -> active(0)
   mark(add(X1, X2)) -> active(add(mark(X1), mark(X2)))
   mark(dbl(X)) -> active(dbl(mark(X)))
   mark(first(X1, X2)) -> active(first(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   terms(mark(X)) -> terms(X)
   terms(active(X)) -> terms(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)
   recip(mark(X)) -> recip(X)
   recip(active(X)) -> recip(X)
   sqr(mark(X)) -> sqr(X)
   sqr(active(X)) -> sqr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   add(mark(X1), X2) -> add(X1, X2)
   add(X1, mark(X2)) -> add(X1, X2)
   add(active(X1), X2) -> add(X1, X2)
   add(X1, active(X2)) -> add(X1, X2)
   dbl(mark(X)) -> dbl(X)
   dbl(active(X)) -> dbl(X)
   first(mark(X1), X2) -> first(X1, X2)
   first(X1, mark(X2)) -> first(X1, X2)
   first(active(X1), X2) -> first(X1, X2)
   first(X1, active(X2)) -> first(X1, X2)

The set Q consists of the following terms:

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

(34) 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.
----------------------------------------

(35)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SQR(active(X)) -> SQR(X)
   SQR(mark(X)) -> SQR(X)

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

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

(36) 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].

   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))


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

(37)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SQR(active(X)) -> SQR(X)
   SQR(mark(X)) -> SQR(X)

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

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)

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

(38) 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:
*SQR(active(X)) -> SQR(X)
The graph contains the following edges 1 > 1


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


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

(39)
YES

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

(40)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   RECIP(active(X)) -> RECIP(X)
   RECIP(mark(X)) -> RECIP(X)

The TRS R consists of the following rules:

   active(terms(N)) -> mark(cons(recip(sqr(N)), terms(s(N))))
   active(sqr(0)) -> mark(0)
   active(sqr(s(X))) -> mark(s(add(sqr(X), dbl(X))))
   active(dbl(0)) -> mark(0)
   active(dbl(s(X))) -> mark(s(s(dbl(X))))
   active(add(0, X)) -> mark(X)
   active(add(s(X), Y)) -> mark(s(add(X, Y)))
   active(first(0, X)) -> mark(nil)
   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
   mark(terms(X)) -> active(terms(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(recip(X)) -> active(recip(mark(X)))
   mark(sqr(X)) -> active(sqr(mark(X)))
   mark(s(X)) -> active(s(mark(X)))
   mark(0) -> active(0)
   mark(add(X1, X2)) -> active(add(mark(X1), mark(X2)))
   mark(dbl(X)) -> active(dbl(mark(X)))
   mark(first(X1, X2)) -> active(first(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   terms(mark(X)) -> terms(X)
   terms(active(X)) -> terms(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)
   recip(mark(X)) -> recip(X)
   recip(active(X)) -> recip(X)
   sqr(mark(X)) -> sqr(X)
   sqr(active(X)) -> sqr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   add(mark(X1), X2) -> add(X1, X2)
   add(X1, mark(X2)) -> add(X1, X2)
   add(active(X1), X2) -> add(X1, X2)
   add(X1, active(X2)) -> add(X1, X2)
   dbl(mark(X)) -> dbl(X)
   dbl(active(X)) -> dbl(X)
   first(mark(X1), X2) -> first(X1, X2)
   first(X1, mark(X2)) -> first(X1, X2)
   first(active(X1), X2) -> first(X1, X2)
   first(X1, active(X2)) -> first(X1, X2)

The set Q consists of the following terms:

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

(41) 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.
----------------------------------------

(42)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   RECIP(active(X)) -> RECIP(X)
   RECIP(mark(X)) -> RECIP(X)

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

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

(43) 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].

   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))


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

(44)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   RECIP(active(X)) -> RECIP(X)
   RECIP(mark(X)) -> RECIP(X)

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

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)

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

(45) 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:
*RECIP(active(X)) -> RECIP(X)
The graph contains the following edges 1 > 1


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


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

(46)
YES

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

(47)
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(terms(N)) -> mark(cons(recip(sqr(N)), terms(s(N))))
   active(sqr(0)) -> mark(0)
   active(sqr(s(X))) -> mark(s(add(sqr(X), dbl(X))))
   active(dbl(0)) -> mark(0)
   active(dbl(s(X))) -> mark(s(s(dbl(X))))
   active(add(0, X)) -> mark(X)
   active(add(s(X), Y)) -> mark(s(add(X, Y)))
   active(first(0, X)) -> mark(nil)
   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
   mark(terms(X)) -> active(terms(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(recip(X)) -> active(recip(mark(X)))
   mark(sqr(X)) -> active(sqr(mark(X)))
   mark(s(X)) -> active(s(mark(X)))
   mark(0) -> active(0)
   mark(add(X1, X2)) -> active(add(mark(X1), mark(X2)))
   mark(dbl(X)) -> active(dbl(mark(X)))
   mark(first(X1, X2)) -> active(first(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   terms(mark(X)) -> terms(X)
   terms(active(X)) -> terms(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)
   recip(mark(X)) -> recip(X)
   recip(active(X)) -> recip(X)
   sqr(mark(X)) -> sqr(X)
   sqr(active(X)) -> sqr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   add(mark(X1), X2) -> add(X1, X2)
   add(X1, mark(X2)) -> add(X1, X2)
   add(active(X1), X2) -> add(X1, X2)
   add(X1, active(X2)) -> add(X1, X2)
   dbl(mark(X)) -> dbl(X)
   dbl(active(X)) -> dbl(X)
   first(mark(X1), X2) -> first(X1, X2)
   first(X1, mark(X2)) -> first(X1, X2)
   first(active(X1), X2) -> first(X1, X2)
   first(X1, active(X2)) -> first(X1, X2)

The set Q consists of the following terms:

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

(48) 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.
----------------------------------------

(49)
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(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

(50) 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].

   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))


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

(51)
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(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)

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

(52) 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


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

(53)
YES

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

(54)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TERMS(active(X)) -> TERMS(X)
   TERMS(mark(X)) -> TERMS(X)

The TRS R consists of the following rules:

   active(terms(N)) -> mark(cons(recip(sqr(N)), terms(s(N))))
   active(sqr(0)) -> mark(0)
   active(sqr(s(X))) -> mark(s(add(sqr(X), dbl(X))))
   active(dbl(0)) -> mark(0)
   active(dbl(s(X))) -> mark(s(s(dbl(X))))
   active(add(0, X)) -> mark(X)
   active(add(s(X), Y)) -> mark(s(add(X, Y)))
   active(first(0, X)) -> mark(nil)
   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
   mark(terms(X)) -> active(terms(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(recip(X)) -> active(recip(mark(X)))
   mark(sqr(X)) -> active(sqr(mark(X)))
   mark(s(X)) -> active(s(mark(X)))
   mark(0) -> active(0)
   mark(add(X1, X2)) -> active(add(mark(X1), mark(X2)))
   mark(dbl(X)) -> active(dbl(mark(X)))
   mark(first(X1, X2)) -> active(first(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   terms(mark(X)) -> terms(X)
   terms(active(X)) -> terms(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)
   recip(mark(X)) -> recip(X)
   recip(active(X)) -> recip(X)
   sqr(mark(X)) -> sqr(X)
   sqr(active(X)) -> sqr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   add(mark(X1), X2) -> add(X1, X2)
   add(X1, mark(X2)) -> add(X1, X2)
   add(active(X1), X2) -> add(X1, X2)
   add(X1, active(X2)) -> add(X1, X2)
   dbl(mark(X)) -> dbl(X)
   dbl(active(X)) -> dbl(X)
   first(mark(X1), X2) -> first(X1, X2)
   first(X1, mark(X2)) -> first(X1, X2)
   first(active(X1), X2) -> first(X1, X2)
   first(X1, active(X2)) -> first(X1, X2)

The set Q consists of the following terms:

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

(55) 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.
----------------------------------------

(56)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TERMS(active(X)) -> TERMS(X)
   TERMS(mark(X)) -> TERMS(X)

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

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

(57) 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].

   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))


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

(58)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TERMS(active(X)) -> TERMS(X)
   TERMS(mark(X)) -> TERMS(X)

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

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)

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

(59) 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:
*TERMS(active(X)) -> TERMS(X)
The graph contains the following edges 1 > 1


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


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

(60)
YES

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

(61)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
   ACTIVE(terms(N)) -> MARK(cons(recip(sqr(N)), terms(s(N))))
   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(terms(X)) -> ACTIVE(terms(mark(X)))
   ACTIVE(sqr(s(X))) -> MARK(s(add(sqr(X), dbl(X))))
   MARK(s(X)) -> ACTIVE(s(mark(X)))
   ACTIVE(dbl(s(X))) -> MARK(s(s(dbl(X))))
   MARK(s(X)) -> MARK(X)
   MARK(terms(X)) -> MARK(X)
   MARK(recip(X)) -> ACTIVE(recip(mark(X)))
   ACTIVE(add(0, X)) -> MARK(X)
   MARK(recip(X)) -> MARK(X)
   MARK(sqr(X)) -> ACTIVE(sqr(mark(X)))
   ACTIVE(add(s(X), Y)) -> MARK(s(add(X, Y)))
   ACTIVE(first(s(X), cons(Y, Z))) -> MARK(cons(Y, first(X, Z)))
   MARK(sqr(X)) -> MARK(X)
   MARK(add(X1, X2)) -> ACTIVE(add(mark(X1), mark(X2)))
   MARK(add(X1, X2)) -> MARK(X1)
   MARK(add(X1, X2)) -> MARK(X2)
   MARK(dbl(X)) -> ACTIVE(dbl(mark(X)))
   MARK(dbl(X)) -> MARK(X)
   MARK(first(X1, X2)) -> ACTIVE(first(mark(X1), mark(X2)))
   MARK(first(X1, X2)) -> MARK(X1)
   MARK(first(X1, X2)) -> MARK(X2)

The TRS R consists of the following rules:

   active(terms(N)) -> mark(cons(recip(sqr(N)), terms(s(N))))
   active(sqr(0)) -> mark(0)
   active(sqr(s(X))) -> mark(s(add(sqr(X), dbl(X))))
   active(dbl(0)) -> mark(0)
   active(dbl(s(X))) -> mark(s(s(dbl(X))))
   active(add(0, X)) -> mark(X)
   active(add(s(X), Y)) -> mark(s(add(X, Y)))
   active(first(0, X)) -> mark(nil)
   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
   mark(terms(X)) -> active(terms(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(recip(X)) -> active(recip(mark(X)))
   mark(sqr(X)) -> active(sqr(mark(X)))
   mark(s(X)) -> active(s(mark(X)))
   mark(0) -> active(0)
   mark(add(X1, X2)) -> active(add(mark(X1), mark(X2)))
   mark(dbl(X)) -> active(dbl(mark(X)))
   mark(first(X1, X2)) -> active(first(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   terms(mark(X)) -> terms(X)
   terms(active(X)) -> terms(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)
   recip(mark(X)) -> recip(X)
   recip(active(X)) -> recip(X)
   sqr(mark(X)) -> sqr(X)
   sqr(active(X)) -> sqr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   add(mark(X1), X2) -> add(X1, X2)
   add(X1, mark(X2)) -> add(X1, X2)
   add(active(X1), X2) -> add(X1, X2)
   add(X1, active(X2)) -> add(X1, X2)
   dbl(mark(X)) -> dbl(X)
   dbl(active(X)) -> dbl(X)
   first(mark(X1), X2) -> first(X1, X2)
   first(X1, mark(X2)) -> first(X1, X2)
   first(active(X1), X2) -> first(X1, X2)
   first(X1, active(X2)) -> first(X1, X2)

The set Q consists of the following terms:

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

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


The following pairs can be oriented strictly and are deleted.

   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
   MARK(s(X)) -> ACTIVE(s(mark(X)))
   MARK(recip(X)) -> ACTIVE(recip(mark(X)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( ACTIVE_1(x_1) ) = max{0, 2x_1 - 1}
POL( add_2(x_1, x_2) ) = 1
POL( cons_2(x_1, x_2) ) = max{0, -2}
POL( dbl_1(x_1) ) = 1
POL( first_2(x_1, x_2) ) = 1
POL( recip_1(x_1) ) = max{0, -2}
POL( s_1(x_1) ) = max{0, -2}
POL( sqr_1(x_1) ) = 1
POL( terms_1(x_1) ) = 1
POL( mark_1(x_1) ) = 0
POL( active_1(x_1) ) = 2
POL( 0 ) = 0
POL( nil ) = 0
POL( MARK_1(x_1) ) = 1

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

   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)
   terms(active(X)) -> terms(X)
   terms(mark(X)) -> terms(X)
   s(active(X)) -> s(X)
   s(mark(X)) -> s(X)
   recip(active(X)) -> recip(X)
   recip(mark(X)) -> recip(X)
   sqr(active(X)) -> sqr(X)
   sqr(mark(X)) -> sqr(X)
   add(X1, mark(X2)) -> add(X1, X2)
   add(mark(X1), X2) -> add(X1, X2)
   add(active(X1), X2) -> add(X1, X2)
   add(X1, active(X2)) -> add(X1, X2)
   dbl(active(X)) -> dbl(X)
   dbl(mark(X)) -> dbl(X)
   first(X1, mark(X2)) -> first(X1, X2)
   first(mark(X1), X2) -> first(X1, X2)
   first(active(X1), X2) -> first(X1, X2)
   first(X1, active(X2)) -> first(X1, X2)


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

(63)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVE(terms(N)) -> MARK(cons(recip(sqr(N)), terms(s(N))))
   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(terms(X)) -> ACTIVE(terms(mark(X)))
   ACTIVE(sqr(s(X))) -> MARK(s(add(sqr(X), dbl(X))))
   ACTIVE(dbl(s(X))) -> MARK(s(s(dbl(X))))
   MARK(s(X)) -> MARK(X)
   MARK(terms(X)) -> MARK(X)
   ACTIVE(add(0, X)) -> MARK(X)
   MARK(recip(X)) -> MARK(X)
   MARK(sqr(X)) -> ACTIVE(sqr(mark(X)))
   ACTIVE(add(s(X), Y)) -> MARK(s(add(X, Y)))
   ACTIVE(first(s(X), cons(Y, Z))) -> MARK(cons(Y, first(X, Z)))
   MARK(sqr(X)) -> MARK(X)
   MARK(add(X1, X2)) -> ACTIVE(add(mark(X1), mark(X2)))
   MARK(add(X1, X2)) -> MARK(X1)
   MARK(add(X1, X2)) -> MARK(X2)
   MARK(dbl(X)) -> ACTIVE(dbl(mark(X)))
   MARK(dbl(X)) -> MARK(X)
   MARK(first(X1, X2)) -> ACTIVE(first(mark(X1), mark(X2)))
   MARK(first(X1, X2)) -> MARK(X1)
   MARK(first(X1, X2)) -> MARK(X2)

The TRS R consists of the following rules:

   active(terms(N)) -> mark(cons(recip(sqr(N)), terms(s(N))))
   active(sqr(0)) -> mark(0)
   active(sqr(s(X))) -> mark(s(add(sqr(X), dbl(X))))
   active(dbl(0)) -> mark(0)
   active(dbl(s(X))) -> mark(s(s(dbl(X))))
   active(add(0, X)) -> mark(X)
   active(add(s(X), Y)) -> mark(s(add(X, Y)))
   active(first(0, X)) -> mark(nil)
   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
   mark(terms(X)) -> active(terms(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(recip(X)) -> active(recip(mark(X)))
   mark(sqr(X)) -> active(sqr(mark(X)))
   mark(s(X)) -> active(s(mark(X)))
   mark(0) -> active(0)
   mark(add(X1, X2)) -> active(add(mark(X1), mark(X2)))
   mark(dbl(X)) -> active(dbl(mark(X)))
   mark(first(X1, X2)) -> active(first(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   terms(mark(X)) -> terms(X)
   terms(active(X)) -> terms(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)
   recip(mark(X)) -> recip(X)
   recip(active(X)) -> recip(X)
   sqr(mark(X)) -> sqr(X)
   sqr(active(X)) -> sqr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   add(mark(X1), X2) -> add(X1, X2)
   add(X1, mark(X2)) -> add(X1, X2)
   add(active(X1), X2) -> add(X1, X2)
   add(X1, active(X2)) -> add(X1, X2)
   dbl(mark(X)) -> dbl(X)
   dbl(active(X)) -> dbl(X)
   first(mark(X1), X2) -> first(X1, X2)
   first(X1, mark(X2)) -> first(X1, X2)
   first(active(X1), X2) -> first(X1, X2)
   first(X1, active(X2)) -> first(X1, X2)

The set Q consists of the following terms:

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

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


The following pairs can be oriented strictly and are deleted.

   ACTIVE(terms(N)) -> MARK(cons(recip(sqr(N)), terms(s(N))))
   MARK(terms(X)) -> ACTIVE(terms(mark(X)))
   ACTIVE(sqr(s(X))) -> MARK(s(add(sqr(X), dbl(X))))
   ACTIVE(dbl(s(X))) -> MARK(s(s(dbl(X))))
   MARK(s(X)) -> MARK(X)
   MARK(terms(X)) -> MARK(X)
   ACTIVE(add(0, X)) -> MARK(X)
   MARK(recip(X)) -> MARK(X)
   MARK(sqr(X)) -> ACTIVE(sqr(mark(X)))
   ACTIVE(add(s(X), Y)) -> MARK(s(add(X, Y)))
   ACTIVE(first(s(X), cons(Y, Z))) -> MARK(cons(Y, first(X, Z)))
   MARK(sqr(X)) -> MARK(X)
   MARK(add(X1, X2)) -> ACTIVE(add(mark(X1), mark(X2)))
   MARK(add(X1, X2)) -> MARK(X1)
   MARK(add(X1, X2)) -> MARK(X2)
   MARK(dbl(X)) -> ACTIVE(dbl(mark(X)))
   MARK(dbl(X)) -> MARK(X)
   MARK(first(X1, X2)) -> ACTIVE(first(mark(X1), mark(X2)))
   MARK(first(X1, X2)) -> MARK(X1)
   MARK(first(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

terms(x1)  =  terms(x1)

MARK(x1)  =  MARK(x1)

cons(x1, x2)  =  x1

recip(x1)  =  recip(x1)

sqr(x1)  =  sqr(x1)

s(x1)  =  s(x1)

mark(x1)  =  x1

add(x1, x2)  =  add(x1, x2)

dbl(x1)  =  dbl(x1)

0  =  0

first(x1, x2)  =  first(x1, x2)

active(x1)  =  x1

nil  =  nil


Recursive path order with status [RPO].
Quasi-Precedence: terms_1 > sqr_1 > add_2 > s_1 > [MARK_1, recip_1, 0] > nil
terms_1 > sqr_1 > dbl_1 > s_1 > [MARK_1, recip_1, 0] > nil
first_2 > [MARK_1, recip_1, 0] > nil

Status: terms_1: [1]
MARK_1: multiset status
recip_1: multiset status
sqr_1: [1]
s_1: [1]
add_2: multiset status
dbl_1: [1]
0: multiset status
first_2: [1,2]
nil: multiset status


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

   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   active(terms(N)) -> mark(cons(recip(sqr(N)), terms(s(N))))
   active(sqr(s(X))) -> mark(s(add(sqr(X), dbl(X))))
   mark(s(X)) -> active(s(mark(X)))
   active(dbl(s(X))) -> mark(s(s(dbl(X))))
   active(add(0, X)) -> mark(X)
   mark(terms(X)) -> active(terms(mark(X)))
   active(add(s(X), Y)) -> mark(s(add(X, Y)))
   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
   mark(recip(X)) -> active(recip(mark(X)))
   mark(sqr(X)) -> active(sqr(mark(X)))
   mark(add(X1, X2)) -> active(add(mark(X1), mark(X2)))
   mark(dbl(X)) -> active(dbl(mark(X)))
   mark(first(X1, X2)) -> active(first(mark(X1), mark(X2)))
   mark(0) -> active(0)
   mark(nil) -> active(nil)
   terms(active(X)) -> terms(X)
   terms(mark(X)) -> terms(X)
   sqr(active(X)) -> sqr(X)
   sqr(mark(X)) -> sqr(X)
   add(X1, mark(X2)) -> add(X1, X2)
   add(mark(X1), X2) -> add(X1, X2)
   add(active(X1), X2) -> add(X1, X2)
   add(X1, active(X2)) -> add(X1, X2)
   dbl(active(X)) -> dbl(X)
   dbl(mark(X)) -> dbl(X)
   first(X1, mark(X2)) -> first(X1, X2)
   first(mark(X1), X2) -> first(X1, X2)
   first(active(X1), X2) -> first(X1, X2)
   first(X1, active(X2)) -> first(X1, X2)
   active(sqr(0)) -> mark(0)
   active(dbl(0)) -> mark(0)
   active(first(0, X)) -> mark(nil)
   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)
   recip(active(X)) -> recip(X)
   recip(mark(X)) -> recip(X)


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

(65)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   active(terms(N)) -> mark(cons(recip(sqr(N)), terms(s(N))))
   active(sqr(0)) -> mark(0)
   active(sqr(s(X))) -> mark(s(add(sqr(X), dbl(X))))
   active(dbl(0)) -> mark(0)
   active(dbl(s(X))) -> mark(s(s(dbl(X))))
   active(add(0, X)) -> mark(X)
   active(add(s(X), Y)) -> mark(s(add(X, Y)))
   active(first(0, X)) -> mark(nil)
   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
   mark(terms(X)) -> active(terms(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(recip(X)) -> active(recip(mark(X)))
   mark(sqr(X)) -> active(sqr(mark(X)))
   mark(s(X)) -> active(s(mark(X)))
   mark(0) -> active(0)
   mark(add(X1, X2)) -> active(add(mark(X1), mark(X2)))
   mark(dbl(X)) -> active(dbl(mark(X)))
   mark(first(X1, X2)) -> active(first(mark(X1), mark(X2)))
   mark(nil) -> active(nil)
   terms(mark(X)) -> terms(X)
   terms(active(X)) -> terms(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)
   recip(mark(X)) -> recip(X)
   recip(active(X)) -> recip(X)
   sqr(mark(X)) -> sqr(X)
   sqr(active(X)) -> sqr(X)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   add(mark(X1), X2) -> add(X1, X2)
   add(X1, mark(X2)) -> add(X1, X2)
   add(active(X1), X2) -> add(X1, X2)
   add(X1, active(X2)) -> add(X1, X2)
   dbl(mark(X)) -> dbl(X)
   dbl(active(X)) -> dbl(X)
   first(mark(X1), X2) -> first(X1, X2)
   first(X1, mark(X2)) -> first(X1, X2)
   first(active(X1), X2) -> first(X1, X2)
   first(X1, active(X2)) -> first(X1, X2)

The set Q consists of the following terms:

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

(66) 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.
----------------------------------------

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

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

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

   active(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))

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

(68) 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(terms(x0))
   active(sqr(0))
   active(sqr(s(x0)))
   active(dbl(0))
   active(dbl(s(x0)))
   active(add(0, x0))
   active(add(s(x0), x1))
   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   mark(terms(x0))
   mark(cons(x0, x1))
   mark(recip(x0))
   mark(sqr(x0))
   mark(s(x0))
   mark(0)
   mark(add(x0, x1))
   mark(dbl(x0))
   mark(first(x0, x1))
   mark(nil)
   terms(mark(x0))
   terms(active(x0))
   recip(mark(x0))
   recip(active(x0))
   sqr(mark(x0))
   sqr(active(x0))
   s(mark(x0))
   s(active(x0))
   add(mark(x0), x1)
   add(x0, mark(x1))
   add(active(x0), x1)
   add(x0, active(x1))
   dbl(mark(x0))
   dbl(active(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))


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

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

   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))

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

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


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

(71)
YES