/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES
    (21) QDP
        (22) UsableRulesProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) QReductionProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) TransformationProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) TransformationProof [EQUIVALENT, 0 ms]
        (29) QDP
        (30) DependencyGraphProof [EQUIVALENT, 0 ms]
        (31) QDP
        (32) TransformationProof [EQUIVALENT, 0 ms]
        (33) QDP
        (34) TransformationProof [EQUIVALENT, 0 ms]
        (35) QDP
        (36) TransformationProof [EQUIVALENT, 0 ms]
        (37) QDP
        (38) TransformationProof [EQUIVALENT, 0 ms]
        (39) QDP
        (40) TransformationProof [EQUIVALENT, 0 ms]
        (41) QDP
        (42) TransformationProof [EQUIVALENT, 0 ms]
        (43) QDP
        (44) DependencyGraphProof [EQUIVALENT, 0 ms]
        (45) QDP
        (46) TransformationProof [EQUIVALENT, 0 ms]
        (47) QDP
        (48) TransformationProof [EQUIVALENT, 0 ms]
        (49) QDP
        (50) DependencyGraphProof [EQUIVALENT, 0 ms]
        (51) AND
            (52) QDP
                (53) UsableRulesProof [EQUIVALENT, 0 ms]
                (54) QDP
                (55) TransformationProof [EQUIVALENT, 0 ms]
                (56) QDP
                (57) UsableRulesProof [EQUIVALENT, 0 ms]
                (58) QDP
                (59) QReductionProof [EQUIVALENT, 0 ms]
                (60) QDP
                (61) TransformationProof [EQUIVALENT, 0 ms]
                (62) QDP
                (63) UsableRulesProof [EQUIVALENT, 0 ms]
                (64) QDP
                (65) QReductionProof [EQUIVALENT, 0 ms]
                (66) QDP
                (67) TransformationProof [EQUIVALENT, 0 ms]
                (68) QDP
                (69) UsableRulesProof [EQUIVALENT, 0 ms]
                (70) QDP
                (71) QReductionProof [EQUIVALENT, 0 ms]
                (72) QDP
                (73) QDPSizeChangeProof [EQUIVALENT, 0 ms]
                (74) YES
            (75) QDP
                (76) UsableRulesProof [EQUIVALENT, 0 ms]
                (77) QDP
                (78) QReductionProof [EQUIVALENT, 0 ms]
                (79) QDP
                (80) QDPSizeChangeProof [EQUIVALENT, 0 ms]
                (81) 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(gr(add(x, y), 0), p(x), y)
   cond2(false, x, y) -> cond3(gr(y, 0), x, y)
   cond3(true, x, y) -> cond1(gr(add(x, y), 0), x, p(y))
   cond3(false, x, y) -> cond1(gr(add(x, y), 0), x, y)
   gr(0, x) -> false
   gr(s(x), 0) -> true
   gr(s(x), s(y)) -> gr(x, y)
   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   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)
   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   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(gr(add(x, y), 0), p(x), y)
   cond2(false, x, y) -> cond3(gr(y, 0), x, y)
   cond3(true, x, y) -> cond1(gr(add(x, y), 0), x, p(y))
   cond3(false, x, y) -> cond1(gr(add(x, 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(gr(add(x, y), 0), p(x), y)
   cond2(false, x, y) -> cond3(gr(y, 0), x, y)
   cond3(true, x, y) -> cond1(gr(add(x, y), 0), x, p(y))
   cond3(false, x, y) -> cond1(gr(add(x, y), 0), x, y)
   gr(0, x) -> false
   gr(s(x), 0) -> true
   gr(s(x), s(y)) -> gr(x, y)
   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   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))
   add(0, x0)
   add(s(x0), x1)
   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(gr(add(x, y), 0), p(x), y)
   COND2(true, x, y) -> GR(add(x, y), 0)
   COND2(true, x, y) -> ADD(x, y)
   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(gr(add(x, y), 0), x, p(y))
   COND3(true, x, y) -> GR(add(x, y), 0)
   COND3(true, x, y) -> ADD(x, y)
   COND3(true, x, y) -> P(y)
   COND3(false, x, y) -> COND1(gr(add(x, y), 0), x, y)
   COND3(false, x, y) -> GR(add(x, y), 0)
   COND3(false, x, y) -> ADD(x, y)
   GR(s(x), s(y)) -> GR(x, y)
   ADD(s(x), y) -> ADD(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(gr(add(x, y), 0), p(x), y)
   cond2(false, x, y) -> cond3(gr(y, 0), x, y)
   cond3(true, x, y) -> cond1(gr(add(x, y), 0), x, p(y))
   cond3(false, x, y) -> cond1(gr(add(x, y), 0), x, y)
   gr(0, x) -> false
   gr(s(x), 0) -> true
   gr(s(x), s(y)) -> gr(x, y)
   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   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))
   add(0, x0)
   add(s(x0), x1)
   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 3 SCCs with 10 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   ADD(s(x), y) -> ADD(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(gr(add(x, y), 0), p(x), y)
   cond2(false, x, y) -> cond3(gr(y, 0), x, y)
   cond3(true, x, y) -> cond1(gr(add(x, y), 0), x, p(y))
   cond3(false, x, y) -> cond1(gr(add(x, y), 0), x, y)
   gr(0, x) -> false
   gr(s(x), 0) -> true
   gr(s(x), s(y)) -> gr(x, y)
   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   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))
   add(0, x0)
   add(s(x0), x1)
   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:

   ADD(s(x), y) -> ADD(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))
   add(0, x0)
   add(s(x0), x1)
   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))
   add(0, x0)
   add(s(x0), x1)
   p(0)
   p(s(x0))


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

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

   ADD(s(x), y) -> ADD(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:
*ADD(s(x), y) -> ADD(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:

   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(gr(add(x, y), 0), p(x), y)
   cond2(false, x, y) -> cond3(gr(y, 0), x, y)
   cond3(true, x, y) -> cond1(gr(add(x, y), 0), x, p(y))
   cond3(false, x, y) -> cond1(gr(add(x, y), 0), x, y)
   gr(0, x) -> false
   gr(s(x), 0) -> true
   gr(s(x), s(y)) -> gr(x, y)
   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   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))
   add(0, x0)
   add(s(x0), x1)
   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:

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


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

(18)
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.
----------------------------------------

(19) 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


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

(20)
YES

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

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

   COND2(true, x, y) -> COND1(gr(add(x, 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(gr(add(x, y), 0), x, p(y))
   COND3(false, x, y) -> COND1(gr(add(x, 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(gr(add(x, y), 0), p(x), y)
   cond2(false, x, y) -> cond3(gr(y, 0), x, y)
   cond3(true, x, y) -> cond1(gr(add(x, y), 0), x, p(y))
   cond3(false, x, y) -> cond1(gr(add(x, y), 0), x, y)
   gr(0, x) -> false
   gr(s(x), 0) -> true
   gr(s(x), s(y)) -> gr(x, y)
   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   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))
   add(0, x0)
   add(s(x0), x1)
   p(0)
   p(s(x0))

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

(22) 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.
----------------------------------------

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

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

The TRS R consists of the following rules:

   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   gr(0, x) -> false
   gr(s(x), 0) -> 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))
   add(0, x0)
   add(s(x0), x1)
   p(0)
   p(s(x0))

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

(24) 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)


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

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

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

The TRS R consists of the following rules:

   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   gr(0, x) -> false
   gr(s(x), 0) -> 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))
   add(0, x0)
   add(s(x0), x1)
   p(0)
   p(s(x0))

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

(26) 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))


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

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

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

   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   gr(0, x) -> false
   gr(s(x), 0) -> 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))
   add(0, x0)
   add(s(x0), x1)
   p(0)
   p(s(x0))

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

(28) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule COND2(true, x, y) -> COND1(gr(add(x, y), 0), p(x), y) at position [1] we obtained the following new rules [LPAR04]:

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


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

(29)
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(gr(add(x, y), 0), x, p(y))
   COND3(false, x, y) -> COND1(gr(add(x, y), 0), x, y)
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, 0, y1) -> COND1(gr(add(0, y1), 0), 0, y1)
   COND2(true, s(x0), y1) -> COND1(gr(add(s(x0), y1), 0), x0, y1)

The TRS R consists of the following rules:

   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   gr(0, x) -> false
   gr(s(x), 0) -> 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))
   add(0, x0)
   add(s(x0), x1)
   p(0)
   p(s(x0))

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

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

(31)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND3(true, x, y) -> COND1(gr(add(x, 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(gr(add(x, y), 0), x, y)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), y1) -> COND1(gr(add(s(x0), y1), 0), x0, y1)

The TRS R consists of the following rules:

   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   gr(0, x) -> false
   gr(s(x), 0) -> 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))
   add(0, x0)
   add(s(x0), x1)
   p(0)
   p(s(x0))

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

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

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


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

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

   COND3(true, x, y) -> COND1(gr(add(x, 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(gr(add(x, y), 0), x, y)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), y1) -> COND1(gr(s(add(x0, y1)), 0), x0, y1)

The TRS R consists of the following rules:

   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   gr(0, x) -> false
   gr(s(x), 0) -> 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))
   add(0, x0)
   add(s(x0), x1)
   p(0)
   p(s(x0))

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

(34) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule COND2(true, s(x0), y1) -> COND1(gr(s(add(x0, y1)), 0), x0, y1) at position [0] 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))


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

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

   COND3(true, x, y) -> COND1(gr(add(x, 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(gr(add(x, y), 0), x, y)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)

The TRS R consists of the following rules:

   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   gr(0, x) -> false
   gr(s(x), 0) -> 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))
   add(0, x0)
   add(s(x0), x1)
   p(0)
   p(s(x0))

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

(36) 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)))


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

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

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

   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   gr(0, x) -> false
   gr(s(x), 0) -> 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))
   add(0, x0)
   add(s(x0), x1)
   p(0)
   p(s(x0))

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

(38) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule COND3(false, x, y) -> COND1(gr(add(x, y), 0), x, y) we obtained the following new rules [LPAR04]:

   (COND3(false, z0, 0) -> COND1(gr(add(z0, 0), 0), z0, 0),COND3(false, z0, 0) -> COND1(gr(add(z0, 0), 0), z0, 0))


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

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

   COND3(true, x, y) -> COND1(gr(add(x, y), 0), x, p(y))
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND2(false, y0, 0) -> COND3(false, y0, 0)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(false, z0, 0) -> COND1(gr(add(z0, 0), 0), z0, 0)

The TRS R consists of the following rules:

   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   gr(0, x) -> false
   gr(s(x), 0) -> 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))
   add(0, x0)
   add(s(x0), x1)
   p(0)
   p(s(x0))

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

(40) 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))


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

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

   COND3(true, x, y) -> COND1(gr(add(x, y), 0), x, p(y))
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND2(false, y0, s(x0)) -> COND3(true, y0, s(x0))
   COND3(false, z0, 0) -> COND1(gr(add(z0, 0), 0), z0, 0)
   COND2(false, 0, 0) -> COND3(false, 0, 0)

The TRS R consists of the following rules:

   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   gr(0, x) -> false
   gr(s(x), 0) -> 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))
   add(0, x0)
   add(s(x0), x1)
   p(0)
   p(s(x0))

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

(42) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule COND3(false, z0, 0) -> COND1(gr(add(z0, 0), 0), z0, 0) at position [0] we obtained the following new rules [LPAR04]:

   (COND3(false, 0, 0) -> COND1(gr(0, 0), 0, 0),COND3(false, 0, 0) -> COND1(gr(0, 0), 0, 0))
   (COND3(false, s(x0), 0) -> COND1(gr(s(add(x0, 0)), 0), s(x0), 0),COND3(false, s(x0), 0) -> COND1(gr(s(add(x0, 0)), 0), s(x0), 0))


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

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

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

The TRS R consists of the following rules:

   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   gr(0, x) -> false
   gr(s(x), 0) -> 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))
   add(0, x0)
   add(s(x0), x1)
   p(0)
   p(s(x0))

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

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

(45)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   gr(0, x) -> false
   gr(s(x), 0) -> 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))
   add(0, x0)
   add(s(x0), x1)
   p(0)
   p(s(x0))

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

(46) 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)))


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

(47)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND3(true, x, y) -> COND1(gr(add(x, y), 0), x, p(y))
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND2(false, 0, s(x1)) -> COND3(true, 0, s(x1))

The TRS R consists of the following rules:

   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   gr(0, x) -> false
   gr(s(x), 0) -> 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))
   add(0, x0)
   add(s(x0), x1)
   p(0)
   p(s(x0))

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

(48) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule COND3(true, x, y) -> COND1(gr(add(x, y), 0), x, p(y)) we obtained the following new rules [LPAR04]:

   (COND3(true, 0, s(z0)) -> COND1(gr(add(0, s(z0)), 0), 0, p(s(z0))),COND3(true, 0, s(z0)) -> COND1(gr(add(0, s(z0)), 0), 0, p(s(z0))))


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

(49)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND1(true, s(x0), y1) -> COND2(true, s(x0), y1)
   COND2(true, s(x0), y1) -> COND1(true, x0, y1)
   COND2(false, 0, s(x1)) -> COND3(true, 0, s(x1))
   COND3(true, 0, s(z0)) -> COND1(gr(add(0, s(z0)), 0), 0, p(s(z0)))

The TRS R consists of the following rules:

   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   gr(0, x) -> false
   gr(s(x), 0) -> 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))
   add(0, x0)
   add(s(x0), x1)
   p(0)
   p(s(x0))

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

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

(51)
Complex Obligation (AND)

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

(52)
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(z0)) -> COND1(gr(add(0, s(z0)), 0), 0, p(s(z0)))
   COND1(true, 0, y1) -> COND2(false, 0, y1)

The TRS R consists of the following rules:

   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   gr(0, x) -> false
   gr(s(x), 0) -> 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))
   add(0, x0)
   add(s(x0), x1)
   p(0)
   p(s(x0))

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

(53) 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.
----------------------------------------

(54)
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(z0)) -> COND1(gr(add(0, s(z0)), 0), 0, p(s(z0)))
   COND1(true, 0, y1) -> COND2(false, 0, y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(55) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule COND3(true, 0, s(z0)) -> COND1(gr(add(0, s(z0)), 0), 0, p(s(z0))) at position [0,0] we obtained the following new rules [LPAR04]:

   (COND3(true, 0, s(z0)) -> COND1(gr(s(z0), 0), 0, p(s(z0))),COND3(true, 0, s(z0)) -> COND1(gr(s(z0), 0), 0, p(s(z0))))


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

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

   COND2(false, 0, s(x1)) -> COND3(true, 0, s(x1))
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND3(true, 0, s(z0)) -> COND1(gr(s(z0), 0), 0, p(s(z0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(57) 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.
----------------------------------------

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

   COND2(false, 0, s(x1)) -> COND3(true, 0, s(x1))
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND3(true, 0, s(z0)) -> COND1(gr(s(z0), 0), 0, p(s(z0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(59) 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].

   add(0, x0)
   add(s(x0), x1)


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

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

   COND2(false, 0, s(x1)) -> COND3(true, 0, s(x1))
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND3(true, 0, s(z0)) -> COND1(gr(s(z0), 0), 0, p(s(z0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(61) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule COND3(true, 0, s(z0)) -> COND1(gr(s(z0), 0), 0, p(s(z0))) at position [0] we obtained the following new rules [LPAR04]:

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


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

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

   COND2(false, 0, s(x1)) -> COND3(true, 0, s(x1))
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND3(true, 0, s(z0)) -> COND1(true, 0, p(s(z0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(63) 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.
----------------------------------------

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

   COND2(false, 0, s(x1)) -> COND3(true, 0, s(x1))
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND3(true, 0, s(z0)) -> COND1(true, 0, p(s(z0)))

The TRS R consists of the following rules:

   p(s(x)) -> x

The set Q consists of the following terms:

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

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

(65) 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))


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

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

   COND2(false, 0, s(x1)) -> COND3(true, 0, s(x1))
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND3(true, 0, s(z0)) -> COND1(true, 0, p(s(z0)))

The TRS R consists of the following rules:

   p(s(x)) -> x

The set Q consists of the following terms:

   p(0)
   p(s(x0))

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

(67) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule COND3(true, 0, s(z0)) -> COND1(true, 0, p(s(z0))) at position [2] we obtained the following new rules [LPAR04]:

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


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

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

   COND2(false, 0, s(x1)) -> COND3(true, 0, s(x1))
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND3(true, 0, s(z0)) -> COND1(true, 0, z0)

The TRS R consists of the following rules:

   p(s(x)) -> x

The set Q consists of the following terms:

   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:

   COND2(false, 0, s(x1)) -> COND3(true, 0, s(x1))
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND3(true, 0, s(z0)) -> COND1(true, 0, z0)

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.
----------------------------------------

(71) 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))


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

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

   COND2(false, 0, s(x1)) -> COND3(true, 0, s(x1))
   COND1(true, 0, y1) -> COND2(false, 0, y1)
   COND3(true, 0, s(z0)) -> COND1(true, 0, z0)

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

(73) 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(z0)) -> COND1(true, 0, z0)
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


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

(74)
YES

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

(75)
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)

The TRS R consists of the following rules:

   add(0, x) -> x
   add(s(x), y) -> s(add(x, y))
   gr(0, x) -> false
   gr(s(x), 0) -> 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))
   add(0, x0)
   add(s(x0), x1)
   p(0)
   p(s(x0))

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

(76) 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.
----------------------------------------

(77)
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)

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

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

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

(78) 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))
   add(0, x0)
   add(s(x0), x1)
   p(0)
   p(s(x0))


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

(79)
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)

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

(80) 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:
*COND2(true, s(x0), y1) -> COND1(true, x0, y1)
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3


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


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

(81)
YES