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


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


YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) UsableRulesProof [EQUIVALENT, 0 ms]
        (12) QDP
        (13) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (14) YES
    (15) QDP
        (16) MNOCProof [EQUIVALENT, 0 ms]
        (17) QDP
        (18) UsableRulesProof [EQUIVALENT, 0 ms]
        (19) QDP
        (20) QReductionProof [EQUIVALENT, 0 ms]
        (21) QDP
        (22) TransformationProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) DependencyGraphProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) TransformationProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) DependencyGraphProof [EQUIVALENT, 0 ms]
        (29) QDP
        (30) UsableRulesProof [EQUIVALENT, 0 ms]
        (31) QDP
        (32) QReductionProof [EQUIVALENT, 0 ms]
        (33) QDP
        (34) TransformationProof [EQUIVALENT, 0 ms]
        (35) QDP
        (36) DependencyGraphProof [EQUIVALENT, 0 ms]
        (37) AND
            (38) QDP
                (39) TransformationProof [EQUIVALENT, 0 ms]
                (40) QDP
                (41) TransformationProof [EQUIVALENT, 0 ms]
                (42) QDP
                (43) QDPSizeChangeProof [EQUIVALENT, 0 ms]
                (44) YES
            (45) QDP
                (46) UsableRulesProof [EQUIVALENT, 0 ms]
                (47) QDP
                (48) QReductionProof [EQUIVALENT, 0 ms]
                (49) QDP
                (50) TransformationProof [EQUIVALENT, 0 ms]
                (51) QDP
                (52) TransformationProof [EQUIVALENT, 0 ms]
                (53) QDP
                (54) QDPSizeChangeProof [EQUIVALENT, 0 ms]
                (55) YES


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

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

   p(0) -> s(s(0))
   p(s(x)) -> x
   p(p(s(x))) -> p(x)
   le(p(s(x)), x) -> le(x, x)
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   minus(x, y) -> if(le(x, y), x, y)
   if(true, x, y) -> 0
   if(false, x, y) -> s(minus(p(x), y))

Q is empty.

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

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

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

   P(p(s(x))) -> P(x)
   LE(p(s(x)), x) -> LE(x, x)
   LE(s(x), s(y)) -> LE(x, y)
   MINUS(x, y) -> IF(le(x, y), x, y)
   MINUS(x, y) -> LE(x, y)
   IF(false, x, y) -> MINUS(p(x), y)
   IF(false, x, y) -> P(x)

The TRS R consists of the following rules:

   p(0) -> s(s(0))
   p(s(x)) -> x
   p(p(s(x))) -> p(x)
   le(p(s(x)), x) -> le(x, x)
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   minus(x, y) -> if(le(x, y), x, y)
   if(true, x, y) -> 0
   if(false, x, y) -> s(minus(p(x), y))

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

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 2 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

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

   LE(s(x), s(y)) -> LE(x, y)
   LE(p(s(x)), x) -> LE(x, x)

The TRS R consists of the following rules:

   p(0) -> s(s(0))
   p(s(x)) -> x
   p(p(s(x))) -> p(x)
   le(p(s(x)), x) -> le(x, x)
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   minus(x, y) -> if(le(x, y), x, y)
   if(true, x, y) -> 0
   if(false, x, y) -> s(minus(p(x), y))

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

(6) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

   LE(s(x), s(y)) -> LE(x, y)
   LE(p(s(x)), x) -> LE(x, x)

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

(8) 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:
*LE(s(x), s(y)) -> LE(x, y)
The graph contains the following edges 1 > 1, 2 > 2


*LE(p(s(x)), x) -> LE(x, x)
The graph contains the following edges 1 > 1, 2 >= 1, 1 > 2, 2 >= 2


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

(9)
YES

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

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   P(p(s(x))) -> P(x)

The TRS R consists of the following rules:

   p(0) -> s(s(0))
   p(s(x)) -> x
   p(p(s(x))) -> p(x)
   le(p(s(x)), x) -> le(x, x)
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   minus(x, y) -> if(le(x, y), x, y)
   if(true, x, y) -> 0
   if(false, x, y) -> s(minus(p(x), y))

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

(11) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

   P(p(s(x))) -> P(x)

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

(13) 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:
*P(p(s(x))) -> P(x)
The graph contains the following edges 1 > 1


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

(14)
YES

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

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

   IF(false, x, y) -> MINUS(p(x), y)
   MINUS(x, y) -> IF(le(x, y), x, y)

The TRS R consists of the following rules:

   p(0) -> s(s(0))
   p(s(x)) -> x
   p(p(s(x))) -> p(x)
   le(p(s(x)), x) -> le(x, x)
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   minus(x, y) -> if(le(x, y), x, y)
   if(true, x, y) -> 0
   if(false, x, y) -> s(minus(p(x), y))

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

(16) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
----------------------------------------

(17)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, x, y) -> MINUS(p(x), y)
   MINUS(x, y) -> IF(le(x, y), x, y)

The TRS R consists of the following rules:

   p(0) -> s(s(0))
   p(s(x)) -> x
   p(p(s(x))) -> p(x)
   le(p(s(x)), x) -> le(x, x)
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   minus(x, y) -> if(le(x, y), x, y)
   if(true, x, y) -> 0
   if(false, x, y) -> s(minus(p(x), y))

The set Q consists of the following terms:

   p(0)
   p(s(x0))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(x0, x1)
   if(true, x0, x1)
   if(false, x0, x1)

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

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

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

   IF(false, x, y) -> MINUS(p(x), y)
   MINUS(x, y) -> IF(le(x, y), x, y)

The TRS R consists of the following rules:

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   p(0) -> s(s(0))
   p(s(x)) -> x

The set Q consists of the following terms:

   p(0)
   p(s(x0))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(x0, x1)
   if(true, x0, x1)
   if(false, x0, x1)

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

(20) 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].

   minus(x0, x1)
   if(true, x0, x1)
   if(false, x0, x1)


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

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

   IF(false, x, y) -> MINUS(p(x), y)
   MINUS(x, y) -> IF(le(x, y), x, y)

The TRS R consists of the following rules:

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   p(0) -> s(s(0))
   p(s(x)) -> x

The set Q consists of the following terms:

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

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

(22) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule MINUS(x, y) -> IF(le(x, y), x, y) at position [0] we obtained the following new rules [LPAR04]:

   (MINUS(0, x0) -> IF(true, 0, x0),MINUS(0, x0) -> IF(true, 0, x0))
   (MINUS(s(x0), 0) -> IF(false, s(x0), 0),MINUS(s(x0), 0) -> IF(false, s(x0), 0))
   (MINUS(s(x0), s(x1)) -> IF(le(x0, x1), s(x0), s(x1)),MINUS(s(x0), s(x1)) -> IF(le(x0, x1), s(x0), s(x1)))


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

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

   IF(false, x, y) -> MINUS(p(x), y)
   MINUS(0, x0) -> IF(true, 0, x0)
   MINUS(s(x0), 0) -> IF(false, s(x0), 0)
   MINUS(s(x0), s(x1)) -> IF(le(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   p(0) -> s(s(0))
   p(s(x)) -> x

The set Q consists of the following terms:

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

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

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

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

   MINUS(s(x0), 0) -> IF(false, s(x0), 0)
   IF(false, x, y) -> MINUS(p(x), y)
   MINUS(s(x0), s(x1)) -> IF(le(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   p(0) -> s(s(0))
   p(s(x)) -> x

The set Q consists of the following terms:

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

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

(26) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule IF(false, x, y) -> MINUS(p(x), y) at position [0] we obtained the following new rules [LPAR04]:

   (IF(false, 0, y1) -> MINUS(s(s(0)), y1),IF(false, 0, y1) -> MINUS(s(s(0)), y1))
   (IF(false, s(x0), y1) -> MINUS(x0, y1),IF(false, s(x0), y1) -> MINUS(x0, y1))


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

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

   MINUS(s(x0), 0) -> IF(false, s(x0), 0)
   MINUS(s(x0), s(x1)) -> IF(le(x0, x1), s(x0), s(x1))
   IF(false, 0, y1) -> MINUS(s(s(0)), y1)
   IF(false, s(x0), y1) -> MINUS(x0, y1)

The TRS R consists of the following rules:

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   p(0) -> s(s(0))
   p(s(x)) -> x

The set Q consists of the following terms:

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

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

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

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

   IF(false, s(x0), y1) -> MINUS(x0, y1)
   MINUS(s(x0), 0) -> IF(false, s(x0), 0)
   MINUS(s(x0), s(x1)) -> IF(le(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   p(0) -> s(s(0))
   p(s(x)) -> x

The set Q consists of the following terms:

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

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

(30) 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.
----------------------------------------

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

   IF(false, s(x0), y1) -> MINUS(x0, y1)
   MINUS(s(x0), 0) -> IF(false, s(x0), 0)
   MINUS(s(x0), s(x1)) -> IF(le(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)

The set Q consists of the following terms:

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

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

(32) 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))


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

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

   IF(false, s(x0), y1) -> MINUS(x0, y1)
   MINUS(s(x0), 0) -> IF(false, s(x0), 0)
   MINUS(s(x0), s(x1)) -> IF(le(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))

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

(34) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IF(false, s(x0), y1) -> MINUS(x0, y1) we obtained the following new rules [LPAR04]:

   (IF(false, s(z0), 0) -> MINUS(z0, 0),IF(false, s(z0), 0) -> MINUS(z0, 0))
   (IF(false, s(z0), s(z1)) -> MINUS(z0, s(z1)),IF(false, s(z0), s(z1)) -> MINUS(z0, s(z1)))


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

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

   MINUS(s(x0), 0) -> IF(false, s(x0), 0)
   MINUS(s(x0), s(x1)) -> IF(le(x0, x1), s(x0), s(x1))
   IF(false, s(z0), 0) -> MINUS(z0, 0)
   IF(false, s(z0), s(z1)) -> MINUS(z0, s(z1))

The TRS R consists of the following rules:

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))

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

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

(37)
Complex Obligation (AND)

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

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

   IF(false, s(z0), s(z1)) -> MINUS(z0, s(z1))
   MINUS(s(x0), s(x1)) -> IF(le(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))

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

(39) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule IF(false, s(z0), s(z1)) -> MINUS(z0, s(z1)) we obtained the following new rules [LPAR04]:

   (IF(false, s(s(y_0)), s(x1)) -> MINUS(s(y_0), s(x1)),IF(false, s(s(y_0)), s(x1)) -> MINUS(s(y_0), s(x1)))


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

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

   MINUS(s(x0), s(x1)) -> IF(le(x0, x1), s(x0), s(x1))
   IF(false, s(s(y_0)), s(x1)) -> MINUS(s(y_0), s(x1))

The TRS R consists of the following rules:

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))

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

(41) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule MINUS(s(x0), s(x1)) -> IF(le(x0, x1), s(x0), s(x1)) we obtained the following new rules [LPAR04]:

   (MINUS(s(s(y_1)), s(x1)) -> IF(le(s(y_1), x1), s(s(y_1)), s(x1)),MINUS(s(s(y_1)), s(x1)) -> IF(le(s(y_1), x1), s(s(y_1)), s(x1)))


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

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

   IF(false, s(s(y_0)), s(x1)) -> MINUS(s(y_0), s(x1))
   MINUS(s(s(y_1)), s(x1)) -> IF(le(s(y_1), x1), s(s(y_1)), s(x1))

The TRS R consists of the following rules:

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))

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

(43) 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:
*MINUS(s(s(y_1)), s(x1)) -> IF(le(s(y_1), x1), s(s(y_1)), s(x1))
The graph contains the following edges 1 >= 2, 2 >= 3


*IF(false, s(s(y_0)), s(x1)) -> MINUS(s(y_0), s(x1))
The graph contains the following edges 2 > 1, 3 >= 2


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

(44)
YES

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

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

   IF(false, s(z0), 0) -> MINUS(z0, 0)
   MINUS(s(x0), 0) -> IF(false, s(x0), 0)

The TRS R consists of the following rules:

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
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(46) 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.
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(47)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, s(z0), 0) -> MINUS(z0, 0)
   MINUS(s(x0), 0) -> IF(false, s(x0), 0)

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

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
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(48) 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].

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))


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

   IF(false, s(z0), 0) -> MINUS(z0, 0)
   MINUS(s(x0), 0) -> IF(false, s(x0), 0)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(50) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule IF(false, s(z0), 0) -> MINUS(z0, 0) we obtained the following new rules [LPAR04]:

   (IF(false, s(s(y_0)), 0) -> MINUS(s(y_0), 0),IF(false, s(s(y_0)), 0) -> MINUS(s(y_0), 0))


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(51)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MINUS(s(x0), 0) -> IF(false, s(x0), 0)
   IF(false, s(s(y_0)), 0) -> MINUS(s(y_0), 0)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(52) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule MINUS(s(x0), 0) -> IF(false, s(x0), 0) we obtained the following new rules [LPAR04]:

   (MINUS(s(s(y_0)), 0) -> IF(false, s(s(y_0)), 0),MINUS(s(s(y_0)), 0) -> IF(false, s(s(y_0)), 0))


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

   IF(false, s(s(y_0)), 0) -> MINUS(s(y_0), 0)
   MINUS(s(s(y_0)), 0) -> IF(false, s(s(y_0)), 0)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(54) 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:
*MINUS(s(s(y_0)), 0) -> IF(false, s(s(y_0)), 0)
The graph contains the following edges 1 >= 2, 2 >= 3


*IF(false, s(s(y_0)), 0) -> MINUS(s(y_0), 0)
The graph contains the following edges 2 > 1, 3 >= 2


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(55)
YES