/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) AAECC Innermost [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 2 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 4 ms]
        (18) QDP
        (19) TransformationProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) TransformationProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) TransformationProof [EQUIVALENT, 0 ms]
        (24) QDP
        (25) TransformationProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) TransformationProof [EQUIVALENT, 0 ms]
        (28) QDP
        (29) TransformationProof [EQUIVALENT, 0 ms]
        (30) QDP
        (31) DependencyGraphProof [EQUIVALENT, 0 ms]
        (32) QDP
        (33) TransformationProof [EQUIVALENT, 0 ms]
        (34) QDP
        (35) TransformationProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) TransformationProof [EQUIVALENT, 0 ms]
        (38) QDP
        (39) TransformationProof [EQUIVALENT, 0 ms]
        (40) QDP
        (41) TransformationProof [EQUIVALENT, 0 ms]
        (42) QDP
        (43) TransformationProof [EQUIVALENT, 0 ms]
        (44) QDP
        (45) DependencyGraphProof [EQUIVALENT, 0 ms]
        (46) QDP
        (47) TransformationProof [EQUIVALENT, 0 ms]
        (48) QDP
        (49) DependencyGraphProof [EQUIVALENT, 0 ms]
        (50) QDP
        (51) TransformationProof [EQUIVALENT, 0 ms]
        (52) QDP
        (53) TransformationProof [EQUIVALENT, 0 ms]
        (54) QDP
        (55) DependencyGraphProof [EQUIVALENT, 0 ms]
        (56) QDP
        (57) TransformationProof [EQUIVALENT, 0 ms]
        (58) QDP
        (59) DependencyGraphProof [EQUIVALENT, 0 ms]
        (60) QDP
        (61) TransformationProof [EQUIVALENT, 0 ms]
        (62) QDP
        (63) TransformationProof [EQUIVALENT, 0 ms]
        (64) QDP
        (65) DependencyGraphProof [EQUIVALENT, 0 ms]
        (66) QDP
        (67) TransformationProof [EQUIVALENT, 0 ms]
        (68) QDP
        (69) UsableRulesProof [EQUIVALENT, 0 ms]
        (70) QDP
        (71) TransformationProof [EQUIVALENT, 0 ms]
        (72) QDP
        (73) TransformationProof [EQUIVALENT, 0 ms]
        (74) QDP
        (75) DependencyGraphProof [EQUIVALENT, 0 ms]
        (76) QDP
        (77) TransformationProof [EQUIVALENT, 0 ms]
        (78) QDP
        (79) DependencyGraphProof [EQUIVALENT, 0 ms]
        (80) QDP
        (81) UsableRulesProof [EQUIVALENT, 0 ms]
        (82) QDP
        (83) QReductionProof [EQUIVALENT, 0 ms]
        (84) QDP
        (85) TransformationProof [EQUIVALENT, 0 ms]
        (86) QDP
        (87) UsableRulesProof [EQUIVALENT, 0 ms]
        (88) QDP
        (89) QReductionProof [EQUIVALENT, 0 ms]
        (90) QDP
        (91) TransformationProof [EQUIVALENT, 0 ms]
        (92) QDP
        (93) TransformationProof [EQUIVALENT, 0 ms]
        (94) QDP
        (95) DependencyGraphProof [EQUIVALENT, 0 ms]
        (96) QDP
        (97) TransformationProof [EQUIVALENT, 0 ms]
        (98) QDP
        (99) DependencyGraphProof [EQUIVALENT, 0 ms]
        (100) QDP
        (101) UsableRulesProof [EQUIVALENT, 0 ms]
        (102) QDP
        (103) QReductionProof [EQUIVALENT, 0 ms]
        (104) QDP
        (105) TransformationProof [EQUIVALENT, 0 ms]
        (106) QDP
        (107) DependencyGraphProof [EQUIVALENT, 0 ms]
        (108) AND
            (109) QDP
                (110) QDPSizeChangeProof [EQUIVALENT, 0 ms]
                (111) YES
            (112) QDP
                (113) QDPSizeChangeProof [EQUIVALENT, 0 ms]
                (114) YES


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

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

   cond1(true, x, y) -> cond2(gr(x, 0), x, y)
   cond2(true, x, y) -> cond1(or(gr(x, 0), gr(y, 0)), p(x), y)
   cond2(false, x, y) -> cond3(gr(y, 0), x, y)
   cond3(true, x, y) -> cond1(or(gr(x, 0), gr(y, 0)), x, p(y))
   cond3(false, x, y) -> cond1(or(gr(x, 0), gr(y, 0)), x, y)
   gr(0, x) -> false
   gr(s(x), 0) -> true
   gr(s(x), s(y)) -> gr(x, y)
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

Q is empty.

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

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   gr(0, x) -> false
   gr(s(x), 0) -> true
   gr(s(x), s(y)) -> gr(x, y)
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The TRS R 2 is 
   cond1(true, x, y) -> cond2(gr(x, 0), x, y)
   cond2(true, x, y) -> cond1(or(gr(x, 0), gr(y, 0)), p(x), y)
   cond2(false, x, y) -> cond3(gr(y, 0), x, y)
   cond3(true, x, y) -> cond1(or(gr(x, 0), gr(y, 0)), x, p(y))
   cond3(false, x, y) -> cond1(or(gr(x, 0), gr(y, 0)), x, y)

The signature Sigma is {cond1_3, cond2_3, cond3_3}
----------------------------------------

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

   cond1(true, x, y) -> cond2(gr(x, 0), x, y)
   cond2(true, x, y) -> cond1(or(gr(x, 0), gr(y, 0)), p(x), y)
   cond2(false, x, y) -> cond3(gr(y, 0), x, y)
   cond3(true, x, y) -> cond1(or(gr(x, 0), gr(y, 0)), x, p(y))
   cond3(false, x, y) -> cond1(or(gr(x, 0), gr(y, 0)), x, y)
   gr(0, x) -> false
   gr(s(x), 0) -> true
   gr(s(x), s(y)) -> gr(x, y)
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   cond1(true, x0, x1)
   cond2(true, x0, x1)
   cond2(false, x0, x1)
   cond3(true, x0, x1)
   cond3(false, x0, x1)
   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))


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

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

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, x, y) -> COND2(gr(x, 0), x, y)
   COND1(true, x, y) -> GR(x, 0)
   COND2(true, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), p(x), y)
   COND2(true, x, y) -> OR(gr(x, 0), gr(y, 0))
   COND2(true, x, y) -> GR(x, 0)
   COND2(true, x, y) -> GR(y, 0)
   COND2(true, x, y) -> P(x)
   COND2(false, x, y) -> COND3(gr(y, 0), x, y)
   COND2(false, x, y) -> GR(y, 0)
   COND3(true, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
   COND3(true, x, y) -> OR(gr(x, 0), gr(y, 0))
   COND3(true, x, y) -> GR(x, 0)
   COND3(true, x, y) -> GR(y, 0)
   COND3(true, x, y) -> P(y)
   COND3(false, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, y)
   COND3(false, x, y) -> OR(gr(x, 0), gr(y, 0))
   COND3(false, x, y) -> GR(x, 0)
   COND3(false, x, y) -> GR(y, 0)
   GR(s(x), s(y)) -> GR(x, y)

The TRS R consists of the following rules:

   cond1(true, x, y) -> cond2(gr(x, 0), x, y)
   cond2(true, x, y) -> cond1(or(gr(x, 0), gr(y, 0)), p(x), y)
   cond2(false, x, y) -> cond3(gr(y, 0), x, y)
   cond3(true, x, y) -> cond1(or(gr(x, 0), gr(y, 0)), x, p(y))
   cond3(false, x, y) -> cond1(or(gr(x, 0), gr(y, 0)), x, y)
   gr(0, x) -> false
   gr(s(x), 0) -> true
   gr(s(x), s(y)) -> gr(x, y)
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   cond1(true, x0, x1)
   cond2(true, x0, x1)
   cond2(false, x0, x1)
   cond3(true, x0, x1)
   cond3(false, x0, x1)
   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 13 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   GR(s(x), s(y)) -> GR(x, y)

The TRS R consists of the following rules:

   cond1(true, x, y) -> cond2(gr(x, 0), x, y)
   cond2(true, x, y) -> cond1(or(gr(x, 0), gr(y, 0)), p(x), y)
   cond2(false, x, y) -> cond3(gr(y, 0), x, y)
   cond3(true, x, y) -> cond1(or(gr(x, 0), gr(y, 0)), x, p(y))
   cond3(false, x, y) -> cond1(or(gr(x, 0), gr(y, 0)), x, y)
   gr(0, x) -> false
   gr(s(x), 0) -> true
   gr(s(x), s(y)) -> gr(x, y)
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   cond1(true, x0, x1)
   cond2(true, x0, x1)
   cond2(false, x0, x1)
   cond3(true, x0, x1)
   cond3(false, x0, x1)
   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(8) 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.
----------------------------------------

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

   GR(s(x), s(y)) -> GR(x, y)

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

   cond1(true, x0, x1)
   cond2(true, x0, x1)
   cond2(false, x0, x1)
   cond3(true, x0, x1)
   cond3(false, x0, x1)
   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(10) 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].

   cond1(true, x0, x1)
   cond2(true, x0, x1)
   cond2(false, x0, x1)
   cond3(true, x0, x1)
   cond3(false, x0, x1)
   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))


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

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

   GR(s(x), s(y)) -> GR(x, y)

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

(12) 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:
*GR(s(x), s(y)) -> GR(x, y)
The graph contains the following edges 1 > 1, 2 > 2


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

(13)
YES

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

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

   COND2(true, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), p(x), y)
   COND1(true, x, y) -> COND2(gr(x, 0), x, y)
   COND2(false, x, y) -> COND3(gr(y, 0), x, y)
   COND3(true, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
   COND3(false, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, y)

The TRS R consists of the following rules:

   cond1(true, x, y) -> cond2(gr(x, 0), x, y)
   cond2(true, x, y) -> cond1(or(gr(x, 0), gr(y, 0)), p(x), y)
   cond2(false, x, y) -> cond3(gr(y, 0), x, y)
   cond3(true, x, y) -> cond1(or(gr(x, 0), gr(y, 0)), x, p(y))
   cond3(false, x, y) -> cond1(or(gr(x, 0), gr(y, 0)), x, y)
   gr(0, x) -> false
   gr(s(x), 0) -> true
   gr(s(x), s(y)) -> gr(x, y)
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   cond1(true, x0, x1)
   cond2(true, x0, x1)
   cond2(false, x0, x1)
   cond3(true, x0, x1)
   cond3(false, x0, x1)
   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(15) 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.
----------------------------------------

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

   COND2(true, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), p(x), y)
   COND1(true, x, y) -> COND2(gr(x, 0), x, y)
   COND2(false, x, y) -> COND3(gr(y, 0), x, y)
   COND3(true, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
   COND3(false, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, y)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   cond1(true, x0, x1)
   cond2(true, x0, x1)
   cond2(false, x0, x1)
   cond3(true, x0, x1)
   cond3(false, x0, x1)
   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(17) 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].

   cond1(true, x0, x1)
   cond2(true, x0, x1)
   cond2(false, x0, x1)
   cond3(true, x0, x1)
   cond3(false, x0, x1)


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

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

   COND2(true, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), p(x), y)
   COND1(true, x, y) -> COND2(gr(x, 0), x, y)
   COND2(false, x, y) -> COND3(gr(y, 0), x, y)
   COND3(true, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
   COND3(false, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, y)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(19) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule COND2(true, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), p(x), y) at position [0] we obtained the following new rules [LPAR04]:

   (COND2(true, 0, y1) -> COND1(or(false, gr(y1, 0)), p(0), y1),COND2(true, 0, y1) -> COND1(or(false, gr(y1, 0)), p(0), y1))
   (COND2(true, s(x0), y1) -> COND1(or(true, gr(y1, 0)), p(s(x0)), y1),COND2(true, s(x0), y1) -> COND1(or(true, gr(y1, 0)), p(s(x0)), y1))
   (COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0),COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0))
   (COND2(true, y0, s(x0)) -> COND1(or(gr(y0, 0), true), p(y0), s(x0)),COND2(true, y0, s(x0)) -> COND1(or(gr(y0, 0), true), p(y0), s(x0)))


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

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

   COND1(true, x, y) -> COND2(gr(x, 0), x, y)
   COND2(false, x, y) -> COND3(gr(y, 0), x, y)
   COND3(true, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
   COND3(false, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, y)
   COND2(true, 0, y1) -> COND1(or(false, gr(y1, 0)), p(0), y1)
   COND2(true, s(x0), y1) -> COND1(or(true, gr(y1, 0)), p(s(x0)), y1)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND2(true, y0, s(x0)) -> COND1(or(gr(y0, 0), true), p(y0), s(x0))

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(21) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule COND2(true, 0, y1) -> COND1(or(false, gr(y1, 0)), p(0), y1) at position [1] we obtained the following new rules [LPAR04]:

   (COND2(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, y1),COND2(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, y1))


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

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

   COND1(true, x, y) -> COND2(gr(x, 0), x, y)
   COND2(false, x, y) -> COND3(gr(y, 0), x, y)
   COND3(true, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
   COND3(false, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, y)
   COND2(true, s(x0), y1) -> COND1(or(true, gr(y1, 0)), p(s(x0)), y1)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND2(true, y0, s(x0)) -> COND1(or(gr(y0, 0), true), p(y0), s(x0))
   COND2(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, y1)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(23) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule COND2(true, s(x0), y1) -> COND1(or(true, gr(y1, 0)), p(s(x0)), y1) at position [0] we obtained the following new rules [LPAR04]:

   (COND2(true, s(x0), y1) -> COND1(true, p(s(x0)), y1),COND2(true, s(x0), y1) -> COND1(true, p(s(x0)), y1))


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

(24)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, x, y) -> COND2(gr(x, 0), x, y)
   COND2(false, x, y) -> COND3(gr(y, 0), x, y)
   COND3(true, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
   COND3(false, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, y)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND2(true, y0, s(x0)) -> COND1(or(gr(y0, 0), true), p(y0), s(x0))
   COND2(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, y1)
   COND2(true, s(x0), y1) -> COND1(true, p(s(x0)), y1)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(25) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule COND2(true, y0, s(x0)) -> COND1(or(gr(y0, 0), true), p(y0), s(x0)) at position [0] we obtained the following new rules [LPAR04]:

   (COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0)),COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0)))


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

(26)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, x, y) -> COND2(gr(x, 0), x, y)
   COND2(false, x, y) -> COND3(gr(y, 0), x, y)
   COND3(true, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
   COND3(false, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, y)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND2(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, y1)
   COND2(true, s(x0), y1) -> COND1(true, p(s(x0)), y1)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(27) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule COND2(true, s(x0), y1) -> COND1(true, p(s(x0)), y1) at position [1] we obtained the following new rules [LPAR04]:

   (COND2(true, s(x0), y1) -> COND1(true, x0, y1),COND2(true, s(x0), y1) -> COND1(true, x0, y1))


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

(28)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, x, y) -> COND2(gr(x, 0), x, y)
   COND2(false, x, y) -> COND3(gr(y, 0), x, y)
   COND3(true, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
   COND3(false, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, y)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND2(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, y1)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(29) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule COND1(true, x, y) -> COND2(gr(x, 0), x, y) at position [0] we obtained the following new rules [LPAR04]:

   (COND1(true, 0, y1) -> COND2(false, 0, y1),COND1(true, 0, y1) -> COND2(false, 0, y1))
   (COND1(true, s(x0), y1) -> COND2(true, s(x0), y1),COND1(true, s(x0), y1) -> COND2(true, s(x0), y1))


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

(30)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND2(false, x, y) -> COND3(gr(y, 0), x, y)
   COND3(true, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
   COND3(false, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, y)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND2(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, y1)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

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

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

   COND3(true, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, p(y))
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, x, y) -> COND3(gr(y, 0), x, y)
   COND3(false, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, y)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(33) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule COND3(true, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, p(y)) at position [0] we obtained the following new rules [LPAR04]:

   (COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1)),COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1)))
   (COND3(true, s(x0), y1) -> COND1(or(true, gr(y1, 0)), s(x0), p(y1)),COND3(true, s(x0), y1) -> COND1(or(true, gr(y1, 0)), s(x0), p(y1)))
   (COND3(true, y0, 0) -> COND1(or(gr(y0, 0), false), y0, p(0)),COND3(true, y0, 0) -> COND1(or(gr(y0, 0), false), y0, p(0)))
   (COND3(true, y0, s(x0)) -> COND1(or(gr(y0, 0), true), y0, p(s(x0))),COND3(true, y0, s(x0)) -> COND1(or(gr(y0, 0), true), y0, p(s(x0))))


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

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

   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, x, y) -> COND3(gr(y, 0), x, y)
   COND3(false, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, y)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, s(x0), y1) -> COND1(or(true, gr(y1, 0)), s(x0), p(y1))
   COND3(true, y0, 0) -> COND1(or(gr(y0, 0), false), y0, p(0))
   COND3(true, y0, s(x0)) -> COND1(or(gr(y0, 0), true), y0, p(s(x0)))

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(35) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule COND3(true, s(x0), y1) -> COND1(or(true, gr(y1, 0)), s(x0), p(y1)) at position [0] we obtained the following new rules [LPAR04]:

   (COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1)),COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1)))


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

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

   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, x, y) -> COND3(gr(y, 0), x, y)
   COND3(false, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, y)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, y0, 0) -> COND1(or(gr(y0, 0), false), y0, p(0))
   COND3(true, y0, s(x0)) -> COND1(or(gr(y0, 0), true), y0, p(s(x0)))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(37) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule COND3(true, y0, 0) -> COND1(or(gr(y0, 0), false), y0, p(0)) at position [2] we obtained the following new rules [LPAR04]:

   (COND3(true, y0, 0) -> COND1(or(gr(y0, 0), false), y0, 0),COND3(true, y0, 0) -> COND1(or(gr(y0, 0), false), y0, 0))


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

(38)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, x, y) -> COND3(gr(y, 0), x, y)
   COND3(false, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, y)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, y0, s(x0)) -> COND1(or(gr(y0, 0), true), y0, p(s(x0)))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, 0) -> COND1(or(gr(y0, 0), false), y0, 0)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(39) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule COND3(true, y0, s(x0)) -> COND1(or(gr(y0, 0), true), y0, p(s(x0))) at position [0] we obtained the following new rules [LPAR04]:

   (COND3(true, y0, s(x0)) -> COND1(true, y0, p(s(x0))),COND3(true, y0, s(x0)) -> COND1(true, y0, p(s(x0))))


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

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

   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, x, y) -> COND3(gr(y, 0), x, y)
   COND3(false, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, y)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, 0) -> COND1(or(gr(y0, 0), false), y0, 0)
   COND3(true, y0, s(x0)) -> COND1(true, y0, p(s(x0)))

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(41) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule COND3(true, y0, s(x0)) -> COND1(true, y0, p(s(x0))) at position [2] we obtained the following new rules [LPAR04]:

   (COND3(true, y0, s(x0)) -> COND1(true, y0, x0),COND3(true, y0, s(x0)) -> COND1(true, y0, x0))


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

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

   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, x, y) -> COND3(gr(y, 0), x, y)
   COND3(false, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, y)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, 0) -> COND1(or(gr(y0, 0), false), y0, 0)
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(43) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule COND2(false, x, y) -> COND3(gr(y, 0), x, y) at position [0] we obtained the following new rules [LPAR04]:

   (COND2(false, y0, 0) -> COND3(false, y0, 0),COND2(false, y0, 0) -> COND3(false, y0, 0))
   (COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0)),COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0)))


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

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

   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND3(false, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, y)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, 0) -> COND1(or(gr(y0, 0), false), y0, 0)
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

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

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

   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND3(false, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, y)
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(47) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule COND3(false, x, y) -> COND1(or(gr(x, 0), gr(y, 0)), x, y) at position [0] we obtained the following new rules [LPAR04]:

   (COND3(false, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, y1),COND3(false, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, y1))
   (COND3(false, s(x0), y1) -> COND1(or(true, gr(y1, 0)), s(x0), y1),COND3(false, s(x0), y1) -> COND1(or(true, gr(y1, 0)), s(x0), y1))
   (COND3(false, y0, 0) -> COND1(or(gr(y0, 0), false), y0, 0),COND3(false, y0, 0) -> COND1(or(gr(y0, 0), false), y0, 0))
   (COND3(false, y0, s(x0)) -> COND1(or(gr(y0, 0), true), y0, s(x0)),COND3(false, y0, s(x0)) -> COND1(or(gr(y0, 0), true), y0, s(x0)))


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

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

   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND3(false, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, y1)
   COND3(false, s(x0), y1) -> COND1(or(true, gr(y1, 0)), s(x0), y1)
   COND3(false, y0, 0) -> COND1(or(gr(y0, 0), false), y0, 0)
   COND3(false, y0, s(x0)) -> COND1(or(gr(y0, 0), true), y0, s(x0))

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

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

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

   COND3(false, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, y1)
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND3(false, s(x0), y1) -> COND1(or(true, gr(y1, 0)), s(x0), y1)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND3(false, y0, 0) -> COND1(or(gr(y0, 0), false), y0, 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(51) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule COND3(false, s(x0), y1) -> COND1(or(true, gr(y1, 0)), s(x0), y1) at position [0] we obtained the following new rules [LPAR04]:

   (COND3(false, s(x0), y1) -> COND1(true, s(x0), y1),COND3(false, s(x0), y1) -> COND1(true, s(x0), y1))


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

(52)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND3(false, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, y1)
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND3(false, y0, 0) -> COND1(or(gr(y0, 0), false), y0, 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(53) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule COND3(false, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, y1) at position [0] we obtained the following new rules [LPAR04]:

   (COND3(false, 0, 0) -> COND1(or(false, false), 0, 0),COND3(false, 0, 0) -> COND1(or(false, false), 0, 0))
   (COND3(false, 0, s(x0)) -> COND1(or(false, true), 0, s(x0)),COND3(false, 0, s(x0)) -> COND1(or(false, true), 0, s(x0)))


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

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

   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND3(false, y0, 0) -> COND1(or(gr(y0, 0), false), y0, 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)
   COND3(false, 0, 0) -> COND1(or(false, false), 0, 0)
   COND3(false, 0, s(x0)) -> COND1(or(false, true), 0, s(x0))

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

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

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

   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND3(false, y0, 0) -> COND1(or(gr(y0, 0), false), y0, 0)
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(57) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule COND3(false, y0, 0) -> COND1(or(gr(y0, 0), false), y0, 0) at position [0] we obtained the following new rules [LPAR04]:

   (COND3(false, 0, 0) -> COND1(or(false, false), 0, 0),COND3(false, 0, 0) -> COND1(or(false, false), 0, 0))
   (COND3(false, s(x0), 0) -> COND1(or(true, false), s(x0), 0),COND3(false, s(x0), 0) -> COND1(or(true, false), s(x0), 0))


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

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

   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)
   COND3(false, 0, 0) -> COND1(or(false, false), 0, 0)
   COND3(false, s(x0), 0) -> COND1(or(true, false), s(x0), 0)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

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

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

   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND3(false, s(x0), 0) -> COND1(or(true, false), s(x0), 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(61) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule COND3(false, s(x0), 0) -> COND1(or(true, false), s(x0), 0) at position [0] we obtained the following new rules [LPAR04]:

   (COND3(false, s(x0), 0) -> COND1(true, s(x0), 0),COND3(false, s(x0), 0) -> COND1(true, s(x0), 0))


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

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

   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0)
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND3(false, s(x0), 0) -> COND1(true, s(x0), 0)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(63) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule COND2(true, y0, 0) -> COND1(or(gr(y0, 0), false), p(y0), 0) at position [0] we obtained the following new rules [LPAR04]:

   (COND2(true, 0, 0) -> COND1(or(false, false), p(0), 0),COND2(true, 0, 0) -> COND1(or(false, false), p(0), 0))
   (COND2(true, s(x0), 0) -> COND1(or(true, false), p(s(x0)), 0),COND2(true, s(x0), 0) -> COND1(or(true, false), p(s(x0)), 0))


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

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

   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND3(false, s(x0), 0) -> COND1(true, s(x0), 0)
   COND2(true, 0, 0) -> COND1(or(false, false), p(0), 0)
   COND2(true, s(x0), 0) -> COND1(or(true, false), p(s(x0)), 0)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

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

(66)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)
   COND3(false, s(x0), 0) -> COND1(true, s(x0), 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND2(true, s(x0), 0) -> COND1(or(true, false), p(s(x0)), 0)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(67) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule COND2(true, s(x0), 0) -> COND1(or(true, false), p(s(x0)), 0) at position [0] we obtained the following new rules [LPAR04]:

   (COND2(true, s(x0), 0) -> COND1(true, p(s(x0)), 0),COND2(true, s(x0), 0) -> COND1(true, p(s(x0)), 0))


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

(68)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)
   COND3(false, s(x0), 0) -> COND1(true, s(x0), 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND2(true, s(x0), 0) -> COND1(true, p(s(x0)), 0)

The TRS R consists of the following rules:

   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(true, x) -> true
   or(x, true) -> true
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(69) 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.
----------------------------------------

(70)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)
   COND3(false, s(x0), 0) -> COND1(true, s(x0), 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND2(true, s(x0), 0) -> COND1(true, p(s(x0)), 0)

The TRS R consists of the following rules:

   p(s(x)) -> x
   p(0) -> 0
   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(x, true) -> true

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(71) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule COND2(true, s(x0), 0) -> COND1(true, p(s(x0)), 0) at position [1] we obtained the following new rules [LPAR04]:

   (COND2(true, s(x0), 0) -> COND1(true, x0, 0),COND2(true, s(x0), 0) -> COND1(true, x0, 0))


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

(72)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0))
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)
   COND3(false, s(x0), 0) -> COND1(true, s(x0), 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND2(true, s(x0), 0) -> COND1(true, x0, 0)

The TRS R consists of the following rules:

   p(s(x)) -> x
   p(0) -> 0
   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(x, true) -> true

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(73) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule COND2(true, y0, s(x0)) -> COND1(true, p(y0), s(x0)) at position [1] we obtained the following new rules [LPAR04]:

   (COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1)),COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1)))
   (COND2(true, 0, s(y1)) -> COND1(true, 0, s(y1)),COND2(true, 0, s(y1)) -> COND1(true, 0, s(y1)))


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

(74)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)
   COND3(false, s(x0), 0) -> COND1(true, s(x0), 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND2(true, s(x0), 0) -> COND1(true, x0, 0)
   COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1))
   COND2(true, 0, s(y1)) -> COND1(true, 0, s(y1))

The TRS R consists of the following rules:

   p(s(x)) -> x
   p(0) -> 0
   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(x, true) -> true

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

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

(76)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), 0) -> COND1(true, x0, 0)
   COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1))
   COND3(false, s(x0), 0) -> COND1(true, s(x0), 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)

The TRS R consists of the following rules:

   p(s(x)) -> x
   p(0) -> 0
   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(x, true) -> true

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(77) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule COND3(true, 0, y1) -> COND1(or(false, gr(y1, 0)), 0, p(y1)) at position [0] we obtained the following new rules [LPAR04]:

   (COND3(true, 0, 0) -> COND1(or(false, false), 0, p(0)),COND3(true, 0, 0) -> COND1(or(false, false), 0, p(0)))
   (COND3(true, 0, s(x0)) -> COND1(or(false, true), 0, p(s(x0))),COND3(true, 0, s(x0)) -> COND1(or(false, true), 0, p(s(x0))))


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

(78)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), 0) -> COND1(true, x0, 0)
   COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1))
   COND3(false, s(x0), 0) -> COND1(true, s(x0), 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND3(true, 0, 0) -> COND1(or(false, false), 0, p(0))
   COND3(true, 0, s(x0)) -> COND1(or(false, true), 0, p(s(x0)))

The TRS R consists of the following rules:

   p(s(x)) -> x
   p(0) -> 0
   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(x, true) -> true

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

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

(80)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND2(true, s(x0), 0) -> COND1(true, x0, 0)
   COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1))
   COND3(false, s(x0), 0) -> COND1(true, s(x0), 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND3(true, 0, s(x0)) -> COND1(or(false, true), 0, p(s(x0)))

The TRS R consists of the following rules:

   p(s(x)) -> x
   p(0) -> 0
   gr(0, x) -> false
   gr(s(x), 0) -> true
   or(false, false) -> false
   or(x, true) -> true

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(81) 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.
----------------------------------------

(82)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND2(true, s(x0), 0) -> COND1(true, x0, 0)
   COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1))
   COND3(false, s(x0), 0) -> COND1(true, s(x0), 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND3(true, 0, s(x0)) -> COND1(or(false, true), 0, p(s(x0)))

The TRS R consists of the following rules:

   or(x, true) -> true
   p(s(x)) -> x
   p(0) -> 0

The set Q consists of the following terms:

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))
   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(83) 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].

   gr(0, x0)
   gr(s(x0), 0)
   gr(s(x0), s(x1))


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

(84)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND2(true, s(x0), 0) -> COND1(true, x0, 0)
   COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1))
   COND3(false, s(x0), 0) -> COND1(true, s(x0), 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND3(true, 0, s(x0)) -> COND1(or(false, true), 0, p(s(x0)))

The TRS R consists of the following rules:

   or(x, true) -> true
   p(s(x)) -> x
   p(0) -> 0

The set Q consists of the following terms:

   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(85) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule COND3(true, 0, s(x0)) -> COND1(or(false, true), 0, p(s(x0))) at position [0] we obtained the following new rules [LPAR04]:

   (COND3(true, 0, s(x0)) -> COND1(true, 0, p(s(x0))),COND3(true, 0, s(x0)) -> COND1(true, 0, p(s(x0))))


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

(86)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND2(true, s(x0), 0) -> COND1(true, x0, 0)
   COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1))
   COND3(false, s(x0), 0) -> COND1(true, s(x0), 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND3(true, 0, s(x0)) -> COND1(true, 0, p(s(x0)))

The TRS R consists of the following rules:

   or(x, true) -> true
   p(s(x)) -> x
   p(0) -> 0

The set Q consists of the following terms:

   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(87) 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.
----------------------------------------

(88)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND2(true, s(x0), 0) -> COND1(true, x0, 0)
   COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1))
   COND3(false, s(x0), 0) -> COND1(true, s(x0), 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND3(true, 0, s(x0)) -> COND1(true, 0, p(s(x0)))

The TRS R consists of the following rules:

   p(s(x)) -> x
   p(0) -> 0

The set Q consists of the following terms:

   or(false, false)
   or(true, x0)
   or(x0, true)
   p(0)
   p(s(x0))

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

(89) 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].

   or(false, false)
   or(true, x0)
   or(x0, true)


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

(90)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND2(true, s(x0), 0) -> COND1(true, x0, 0)
   COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1))
   COND3(false, s(x0), 0) -> COND1(true, s(x0), 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND3(true, 0, s(x0)) -> COND1(true, 0, p(s(x0)))

The TRS R consists of the following rules:

   p(s(x)) -> x
   p(0) -> 0

The set Q consists of the following terms:

   p(0)
   p(s(x0))

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

(91) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule COND3(true, 0, s(x0)) -> COND1(true, 0, p(s(x0))) at position [2] we obtained the following new rules [LPAR04]:

   (COND3(true, 0, s(x0)) -> COND1(true, 0, x0),COND3(true, 0, s(x0)) -> COND1(true, 0, x0))


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

(92)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND2(true, s(x0), 0) -> COND1(true, x0, 0)
   COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1))
   COND3(false, s(x0), 0) -> COND1(true, s(x0), 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND3(true, 0, s(x0)) -> COND1(true, 0, x0)

The TRS R consists of the following rules:

   p(s(x)) -> x
   p(0) -> 0

The set Q consists of the following terms:

   p(0)
   p(s(x0))

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

(93) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule COND2(false, y0, 0) -> COND3(false, y0, 0) we obtained the following new rules [LPAR04]:

   (COND2(false, 0, 0) -> COND3(false, 0, 0),COND2(false, 0, 0) -> COND3(false, 0, 0))


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

(94)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND3(false, s(x0), y1) -> COND1(true, s(x0), y1)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND2(true, s(x0), 0) -> COND1(true, x0, 0)
   COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1))
   COND3(false, s(x0), 0) -> COND1(true, s(x0), 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND3(true, 0, s(x0)) -> COND1(true, 0, x0)
   COND2(false, 0, 0) -> COND3(false, 0, 0)

The TRS R consists of the following rules:

   p(s(x)) -> x
   p(0) -> 0

The set Q consists of the following terms:

   p(0)
   p(s(x0))

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

(95) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
----------------------------------------

(96)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(true, s(x0), 0) -> COND1(true, x0, 0)
   COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND3(true, 0, s(x0)) -> COND1(true, 0, x0)

The TRS R consists of the following rules:

   p(s(x)) -> x
   p(0) -> 0

The set Q consists of the following terms:

   p(0)
   p(s(x0))

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

(97) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0)) we obtained the following new rules [LPAR04]:

   (COND2(false, 0, s(x1)) -> COND3(true, 0, s(x1)),COND2(false, 0, s(x1)) -> COND3(true, 0, s(x1)))


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

(98)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND3(true, s(x0), y1) -> COND1(true, s(x0), p(y1))
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(true, s(x0), 0) -> COND1(true, x0, 0)
   COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND3(true, 0, s(x0)) -> COND1(true, 0, x0)
   COND2(false, 0, s(x1)) -> COND3(true, 0, s(x1))

The TRS R consists of the following rules:

   p(s(x)) -> x
   p(0) -> 0

The set Q consists of the following terms:

   p(0)
   p(s(x0))

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

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

(100)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, 0, s(x1)) -> COND3(true, 0, s(x1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), 0) -> COND1(true, x0, 0)
   COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1))
   COND3(true, 0, s(x0)) -> COND1(true, 0, x0)

The TRS R consists of the following rules:

   p(s(x)) -> x
   p(0) -> 0

The set Q consists of the following terms:

   p(0)
   p(s(x0))

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

(101) 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.
----------------------------------------

(102)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, 0, s(x1)) -> COND3(true, 0, s(x1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), 0) -> COND1(true, x0, 0)
   COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1))
   COND3(true, 0, s(x0)) -> COND1(true, 0, x0)

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

   p(0)
   p(s(x0))

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

(103) 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].

   p(0)
   p(s(x0))


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

(104)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, 0, s(x1)) -> COND3(true, 0, s(x1))
   COND3(true, y0, s(x0)) -> COND1(true, y0, x0)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), 0) -> COND1(true, x0, 0)
   COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1))
   COND3(true, 0, s(x0)) -> COND1(true, 0, x0)

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

(105) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule COND3(true, y0, s(x0)) -> COND1(true, y0, x0) we obtained the following new rules [LPAR04]:

   (COND3(true, 0, s(z0)) -> COND1(true, 0, z0),COND3(true, 0, s(x0)) -> COND1(true, 0, x0))


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

(106)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND2(false, 0, s(x1)) -> COND3(true, 0, s(x1))
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), 0) -> COND1(true, x0, 0)
   COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1))
   COND3(true, 0, s(x0)) -> COND1(true, 0, x0)

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

(107) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.
----------------------------------------

(108)
Complex Obligation (AND)

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

(109)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND2(false, 0, s(x1)) -> COND3(true, 0, s(x1))
   COND3(true, 0, s(x0)) -> COND1(true, 0, x0)
   COND1(true, 0, y1) -> COND2(false, 0, y1)

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

(110) 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:
*COND3(true, 0, s(x0)) -> COND1(true, 0, x0)
The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3


*COND1(true, 0, y1) -> COND2(false, 0, y1)
The graph contains the following edges 2 >= 2, 3 >= 3


*COND2(false, 0, s(x1)) -> COND3(true, 0, s(x1))
The graph contains the following edges 2 >= 2, 3 >= 3


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

(111)
YES

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

(112)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND2(true, s(x0), 0) -> COND1(true, x0, 0)
   COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1))

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

(113) 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:
*COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3


*COND2(true, s(x0), y1) -> COND1(true, x0, y1)
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3


*COND2(true, s(x0), 0) -> COND1(true, x0, 0)
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3


*COND2(true, s(x0), s(y1)) -> COND1(true, x0, s(y1))
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3


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

(114)
YES