YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) MNOCProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) MRRProof [EQUIVALENT, 1 ms]
        (13) QDP
        (14) PisEmptyProof [EQUIVALENT, 0 ms]
        (15) YES
    (16) QDP
        (17) MNOCProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) UsableRulesProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QReductionProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (24) YES
    (25) QDP
        (26) MNOCProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) UsableRulesProof [EQUIVALENT, 0 ms]
        (29) QDP
        (30) QReductionProof [EQUIVALENT, 0 ms]
        (31) QDP
        (32) TransformationProof [EQUIVALENT, 0 ms]
        (33) QDP
        (34) DependencyGraphProof [EQUIVALENT, 0 ms]
        (35) QDP
        (36) UsableRulesProof [EQUIVALENT, 0 ms]
        (37) QDP
        (38) QReductionProof [EQUIVALENT, 0 ms]
        (39) QDP
        (40) TransformationProof [EQUIVALENT, 0 ms]
        (41) QDP
        (42) UsableRulesProof [EQUIVALENT, 0 ms]
        (43) QDP
        (44) TransformationProof [EQUIVALENT, 0 ms]
        (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:

   double(x) -> permute(x, x, a)
   permute(x, y, a) -> permute(isZero(x), x, b)
   permute(false, x, b) -> permute(ack(x, x), p(x), c)
   permute(true, x, b) -> 0
   permute(y, x, c) -> s(s(permute(x, y, a)))
   p(0) -> 0
   p(s(x)) -> x
   ack(0, x) -> plus(x, s(0))
   ack(s(x), 0) -> ack(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, s(0)) -> s(x)
   plus(x, 0) -> x
   isZero(0) -> true
   isZero(s(x)) -> false

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:

   DOUBLE(x) -> PERMUTE(x, x, a)
   PERMUTE(x, y, a) -> PERMUTE(isZero(x), x, b)
   PERMUTE(x, y, a) -> ISZERO(x)
   PERMUTE(false, x, b) -> PERMUTE(ack(x, x), p(x), c)
   PERMUTE(false, x, b) -> ACK(x, x)
   PERMUTE(false, x, b) -> P(x)
   PERMUTE(y, x, c) -> PERMUTE(x, y, a)
   ACK(0, x) -> PLUS(x, s(0))
   ACK(s(x), 0) -> ACK(x, s(0))
   ACK(s(x), s(y)) -> ACK(x, ack(s(x), y))
   ACK(s(x), s(y)) -> ACK(s(x), y)
   PLUS(s(x), y) -> PLUS(x, s(y))
   PLUS(x, s(s(y))) -> PLUS(s(x), y)

The TRS R consists of the following rules:

   double(x) -> permute(x, x, a)
   permute(x, y, a) -> permute(isZero(x), x, b)
   permute(false, x, b) -> permute(ack(x, x), p(x), c)
   permute(true, x, b) -> 0
   permute(y, x, c) -> s(s(permute(x, y, a)))
   p(0) -> 0
   p(s(x)) -> x
   ack(0, x) -> plus(x, s(0))
   ack(s(x), 0) -> ack(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, s(0)) -> s(x)
   plus(x, 0) -> x
   isZero(0) -> true
   isZero(s(x)) -> false

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 5 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

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

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

The TRS R consists of the following rules:

   double(x) -> permute(x, x, a)
   permute(x, y, a) -> permute(isZero(x), x, b)
   permute(false, x, b) -> permute(ack(x, x), p(x), c)
   permute(true, x, b) -> 0
   permute(y, x, c) -> s(s(permute(x, y, a)))
   p(0) -> 0
   p(s(x)) -> x
   ack(0, x) -> plus(x, s(0))
   ack(s(x), 0) -> ack(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, s(0)) -> s(x)
   plus(x, 0) -> x
   isZero(0) -> true
   isZero(s(x)) -> false

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

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

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

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

The TRS R consists of the following rules:

   double(x) -> permute(x, x, a)
   permute(x, y, a) -> permute(isZero(x), x, b)
   permute(false, x, b) -> permute(ack(x, x), p(x), c)
   permute(true, x, b) -> 0
   permute(y, x, c) -> s(s(permute(x, y, a)))
   p(0) -> 0
   p(s(x)) -> x
   ack(0, x) -> plus(x, s(0))
   ack(s(x), 0) -> ack(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, s(0)) -> s(x)
   plus(x, 0) -> x
   isZero(0) -> true
   isZero(s(x)) -> false

The set Q consists of the following terms:

   double(x0)
   permute(x0, x1, a)
   permute(false, x0, b)
   permute(true, x0, b)
   permute(x0, x1, c)
   p(0)
   p(s(x0))
   ack(0, x0)
   ack(s(x0), 0)
   ack(s(x0), s(x1))
   plus(0, x0)
   plus(s(x0), x1)
   plus(x0, s(s(x1)))
   plus(x0, s(0))
   plus(x0, 0)
   isZero(0)
   isZero(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:

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

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

   double(x0)
   permute(x0, x1, a)
   permute(false, x0, b)
   permute(true, x0, b)
   permute(x0, x1, c)
   p(0)
   p(s(x0))
   ack(0, x0)
   ack(s(x0), 0)
   ack(s(x0), s(x1))
   plus(0, x0)
   plus(s(x0), x1)
   plus(x0, s(s(x1)))
   plus(x0, s(0))
   plus(x0, 0)
   isZero(0)
   isZero(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].

   double(x0)
   permute(x0, x1, a)
   permute(false, x0, b)
   permute(true, x0, b)
   permute(x0, x1, c)
   p(0)
   p(s(x0))
   ack(0, x0)
   ack(s(x0), 0)
   ack(s(x0), s(x1))
   plus(0, x0)
   plus(s(x0), x1)
   plus(x0, s(s(x1)))
   plus(x0, s(0))
   plus(x0, 0)
   isZero(0)
   isZero(s(x0))


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

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

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

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

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

Strictly oriented dependency pairs:

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


Used ordering: Knuth-Bendix order [KBO] with precedence:s_1 > PLUS_2

and weight map:

   s_1=1
   PLUS_2=0

The variable weight is 1

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

(13)
Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(14) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(15)
YES

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

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

   ACK(s(x), s(y)) -> ACK(x, ack(s(x), y))
   ACK(s(x), 0) -> ACK(x, s(0))
   ACK(s(x), s(y)) -> ACK(s(x), y)

The TRS R consists of the following rules:

   double(x) -> permute(x, x, a)
   permute(x, y, a) -> permute(isZero(x), x, b)
   permute(false, x, b) -> permute(ack(x, x), p(x), c)
   permute(true, x, b) -> 0
   permute(y, x, c) -> s(s(permute(x, y, a)))
   p(0) -> 0
   p(s(x)) -> x
   ack(0, x) -> plus(x, s(0))
   ack(s(x), 0) -> ack(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, s(0)) -> s(x)
   plus(x, 0) -> x
   isZero(0) -> true
   isZero(s(x)) -> false

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

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

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

   ACK(s(x), s(y)) -> ACK(x, ack(s(x), y))
   ACK(s(x), 0) -> ACK(x, s(0))
   ACK(s(x), s(y)) -> ACK(s(x), y)

The TRS R consists of the following rules:

   double(x) -> permute(x, x, a)
   permute(x, y, a) -> permute(isZero(x), x, b)
   permute(false, x, b) -> permute(ack(x, x), p(x), c)
   permute(true, x, b) -> 0
   permute(y, x, c) -> s(s(permute(x, y, a)))
   p(0) -> 0
   p(s(x)) -> x
   ack(0, x) -> plus(x, s(0))
   ack(s(x), 0) -> ack(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, s(0)) -> s(x)
   plus(x, 0) -> x
   isZero(0) -> true
   isZero(s(x)) -> false

The set Q consists of the following terms:

   double(x0)
   permute(x0, x1, a)
   permute(false, x0, b)
   permute(true, x0, b)
   permute(x0, x1, c)
   p(0)
   p(s(x0))
   ack(0, x0)
   ack(s(x0), 0)
   ack(s(x0), s(x1))
   plus(0, x0)
   plus(s(x0), x1)
   plus(x0, s(s(x1)))
   plus(x0, s(0))
   plus(x0, 0)
   isZero(0)
   isZero(s(x0))

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

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

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

   ACK(s(x), s(y)) -> ACK(x, ack(s(x), y))
   ACK(s(x), 0) -> ACK(x, s(0))
   ACK(s(x), s(y)) -> ACK(s(x), y)

The TRS R consists of the following rules:

   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   ack(s(x), 0) -> ack(x, s(0))
   ack(0, x) -> plus(x, s(0))
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(0)) -> s(x)
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, 0) -> x

The set Q consists of the following terms:

   double(x0)
   permute(x0, x1, a)
   permute(false, x0, b)
   permute(true, x0, b)
   permute(x0, x1, c)
   p(0)
   p(s(x0))
   ack(0, x0)
   ack(s(x0), 0)
   ack(s(x0), s(x1))
   plus(0, x0)
   plus(s(x0), x1)
   plus(x0, s(s(x1)))
   plus(x0, s(0))
   plus(x0, 0)
   isZero(0)
   isZero(s(x0))

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

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

   double(x0)
   permute(x0, x1, a)
   permute(false, x0, b)
   permute(true, x0, b)
   permute(x0, x1, c)
   p(0)
   p(s(x0))
   isZero(0)
   isZero(s(x0))


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

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

   ACK(s(x), s(y)) -> ACK(x, ack(s(x), y))
   ACK(s(x), 0) -> ACK(x, s(0))
   ACK(s(x), s(y)) -> ACK(s(x), y)

The TRS R consists of the following rules:

   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   ack(s(x), 0) -> ack(x, s(0))
   ack(0, x) -> plus(x, s(0))
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(0)) -> s(x)
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, 0) -> x

The set Q consists of the following terms:

   ack(0, x0)
   ack(s(x0), 0)
   ack(s(x0), s(x1))
   plus(0, x0)
   plus(s(x0), x1)
   plus(x0, s(s(x1)))
   plus(x0, s(0))
   plus(x0, 0)

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

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

From the DPs we obtained the following set of size-change graphs:
*ACK(s(x), 0) -> ACK(x, s(0))
The graph contains the following edges 1 > 1


*ACK(s(x), s(y)) -> ACK(x, ack(s(x), y))
The graph contains the following edges 1 > 1


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


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

(24)
YES

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

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

   PERMUTE(false, x, b) -> PERMUTE(ack(x, x), p(x), c)
   PERMUTE(y, x, c) -> PERMUTE(x, y, a)
   PERMUTE(x, y, a) -> PERMUTE(isZero(x), x, b)

The TRS R consists of the following rules:

   double(x) -> permute(x, x, a)
   permute(x, y, a) -> permute(isZero(x), x, b)
   permute(false, x, b) -> permute(ack(x, x), p(x), c)
   permute(true, x, b) -> 0
   permute(y, x, c) -> s(s(permute(x, y, a)))
   p(0) -> 0
   p(s(x)) -> x
   ack(0, x) -> plus(x, s(0))
   ack(s(x), 0) -> ack(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, s(0)) -> s(x)
   plus(x, 0) -> x
   isZero(0) -> true
   isZero(s(x)) -> false

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

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

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

   PERMUTE(false, x, b) -> PERMUTE(ack(x, x), p(x), c)
   PERMUTE(y, x, c) -> PERMUTE(x, y, a)
   PERMUTE(x, y, a) -> PERMUTE(isZero(x), x, b)

The TRS R consists of the following rules:

   double(x) -> permute(x, x, a)
   permute(x, y, a) -> permute(isZero(x), x, b)
   permute(false, x, b) -> permute(ack(x, x), p(x), c)
   permute(true, x, b) -> 0
   permute(y, x, c) -> s(s(permute(x, y, a)))
   p(0) -> 0
   p(s(x)) -> x
   ack(0, x) -> plus(x, s(0))
   ack(s(x), 0) -> ack(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, s(0)) -> s(x)
   plus(x, 0) -> x
   isZero(0) -> true
   isZero(s(x)) -> false

The set Q consists of the following terms:

   double(x0)
   permute(x0, x1, a)
   permute(false, x0, b)
   permute(true, x0, b)
   permute(x0, x1, c)
   p(0)
   p(s(x0))
   ack(0, x0)
   ack(s(x0), 0)
   ack(s(x0), s(x1))
   plus(0, x0)
   plus(s(x0), x1)
   plus(x0, s(s(x1)))
   plus(x0, s(0))
   plus(x0, 0)
   isZero(0)
   isZero(s(x0))

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

(28) 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.
----------------------------------------

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

   PERMUTE(false, x, b) -> PERMUTE(ack(x, x), p(x), c)
   PERMUTE(y, x, c) -> PERMUTE(x, y, a)
   PERMUTE(x, y, a) -> PERMUTE(isZero(x), x, b)

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   ack(0, x) -> plus(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   ack(s(x), 0) -> ack(x, s(0))
   p(0) -> 0
   p(s(x)) -> x
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(0)) -> s(x)
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, 0) -> x

The set Q consists of the following terms:

   double(x0)
   permute(x0, x1, a)
   permute(false, x0, b)
   permute(true, x0, b)
   permute(x0, x1, c)
   p(0)
   p(s(x0))
   ack(0, x0)
   ack(s(x0), 0)
   ack(s(x0), s(x1))
   plus(0, x0)
   plus(s(x0), x1)
   plus(x0, s(s(x1)))
   plus(x0, s(0))
   plus(x0, 0)
   isZero(0)
   isZero(s(x0))

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

(30) 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].

   double(x0)
   permute(x0, x1, a)
   permute(false, x0, b)
   permute(true, x0, b)
   permute(x0, x1, c)


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

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

   PERMUTE(false, x, b) -> PERMUTE(ack(x, x), p(x), c)
   PERMUTE(y, x, c) -> PERMUTE(x, y, a)
   PERMUTE(x, y, a) -> PERMUTE(isZero(x), x, b)

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   ack(0, x) -> plus(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   ack(s(x), 0) -> ack(x, s(0))
   p(0) -> 0
   p(s(x)) -> x
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(0)) -> s(x)
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, 0) -> x

The set Q consists of the following terms:

   p(0)
   p(s(x0))
   ack(0, x0)
   ack(s(x0), 0)
   ack(s(x0), s(x1))
   plus(0, x0)
   plus(s(x0), x1)
   plus(x0, s(s(x1)))
   plus(x0, s(0))
   plus(x0, 0)
   isZero(0)
   isZero(s(x0))

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

(32) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule PERMUTE(x, y, a) -> PERMUTE(isZero(x), x, b) at position [0] we obtained the following new rules [LPAR04]:

   (PERMUTE(0, y1, a) -> PERMUTE(true, 0, b),PERMUTE(0, y1, a) -> PERMUTE(true, 0, b))
   (PERMUTE(s(x0), y1, a) -> PERMUTE(false, s(x0), b),PERMUTE(s(x0), y1, a) -> PERMUTE(false, s(x0), b))


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

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

   PERMUTE(false, x, b) -> PERMUTE(ack(x, x), p(x), c)
   PERMUTE(y, x, c) -> PERMUTE(x, y, a)
   PERMUTE(0, y1, a) -> PERMUTE(true, 0, b)
   PERMUTE(s(x0), y1, a) -> PERMUTE(false, s(x0), b)

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   ack(0, x) -> plus(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   ack(s(x), 0) -> ack(x, s(0))
   p(0) -> 0
   p(s(x)) -> x
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(0)) -> s(x)
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, 0) -> x

The set Q consists of the following terms:

   p(0)
   p(s(x0))
   ack(0, x0)
   ack(s(x0), 0)
   ack(s(x0), s(x1))
   plus(0, x0)
   plus(s(x0), x1)
   plus(x0, s(s(x1)))
   plus(x0, s(0))
   plus(x0, 0)
   isZero(0)
   isZero(s(x0))

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

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

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

   PERMUTE(y, x, c) -> PERMUTE(x, y, a)
   PERMUTE(s(x0), y1, a) -> PERMUTE(false, s(x0), b)
   PERMUTE(false, x, b) -> PERMUTE(ack(x, x), p(x), c)

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   ack(0, x) -> plus(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   ack(s(x), 0) -> ack(x, s(0))
   p(0) -> 0
   p(s(x)) -> x
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(0)) -> s(x)
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, 0) -> x

The set Q consists of the following terms:

   p(0)
   p(s(x0))
   ack(0, x0)
   ack(s(x0), 0)
   ack(s(x0), s(x1))
   plus(0, x0)
   plus(s(x0), x1)
   plus(x0, s(s(x1)))
   plus(x0, s(0))
   plus(x0, 0)
   isZero(0)
   isZero(s(x0))

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

(36) 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.
----------------------------------------

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

   PERMUTE(y, x, c) -> PERMUTE(x, y, a)
   PERMUTE(s(x0), y1, a) -> PERMUTE(false, s(x0), b)
   PERMUTE(false, x, b) -> PERMUTE(ack(x, x), p(x), c)

The TRS R consists of the following rules:

   ack(0, x) -> plus(x, s(0))
   ack(s(x), 0) -> ack(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   p(0) -> 0
   p(s(x)) -> x
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(0)) -> s(x)
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, 0) -> x

The set Q consists of the following terms:

   p(0)
   p(s(x0))
   ack(0, x0)
   ack(s(x0), 0)
   ack(s(x0), s(x1))
   plus(0, x0)
   plus(s(x0), x1)
   plus(x0, s(s(x1)))
   plus(x0, s(0))
   plus(x0, 0)
   isZero(0)
   isZero(s(x0))

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

(38) 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].

   isZero(0)
   isZero(s(x0))


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

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

   PERMUTE(y, x, c) -> PERMUTE(x, y, a)
   PERMUTE(s(x0), y1, a) -> PERMUTE(false, s(x0), b)
   PERMUTE(false, x, b) -> PERMUTE(ack(x, x), p(x), c)

The TRS R consists of the following rules:

   ack(0, x) -> plus(x, s(0))
   ack(s(x), 0) -> ack(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   p(0) -> 0
   p(s(x)) -> x
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(0)) -> s(x)
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, 0) -> x

The set Q consists of the following terms:

   p(0)
   p(s(x0))
   ack(0, x0)
   ack(s(x0), 0)
   ack(s(x0), s(x1))
   plus(0, x0)
   plus(s(x0), x1)
   plus(x0, s(s(x1)))
   plus(x0, s(0))
   plus(x0, 0)

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

(40) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule PERMUTE(false, x, b) -> PERMUTE(ack(x, x), p(x), c) we obtained the following new rules [LPAR04]:

   (PERMUTE(false, s(z0), b) -> PERMUTE(ack(s(z0), s(z0)), p(s(z0)), c),PERMUTE(false, s(z0), b) -> PERMUTE(ack(s(z0), s(z0)), p(s(z0)), c))


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

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

   PERMUTE(y, x, c) -> PERMUTE(x, y, a)
   PERMUTE(s(x0), y1, a) -> PERMUTE(false, s(x0), b)
   PERMUTE(false, s(z0), b) -> PERMUTE(ack(s(z0), s(z0)), p(s(z0)), c)

The TRS R consists of the following rules:

   ack(0, x) -> plus(x, s(0))
   ack(s(x), 0) -> ack(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   p(0) -> 0
   p(s(x)) -> x
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(0)) -> s(x)
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, 0) -> x

The set Q consists of the following terms:

   p(0)
   p(s(x0))
   ack(0, x0)
   ack(s(x0), 0)
   ack(s(x0), s(x1))
   plus(0, x0)
   plus(s(x0), x1)
   plus(x0, s(s(x1)))
   plus(x0, s(0))
   plus(x0, 0)

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

(42) 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.
----------------------------------------

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

   PERMUTE(y, x, c) -> PERMUTE(x, y, a)
   PERMUTE(s(x0), y1, a) -> PERMUTE(false, s(x0), b)
   PERMUTE(false, s(z0), b) -> PERMUTE(ack(s(z0), s(z0)), p(s(z0)), c)

The TRS R consists of the following rules:

   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   ack(s(x), 0) -> ack(x, s(0))
   p(s(x)) -> x
   ack(0, x) -> plus(x, s(0))
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(0)) -> s(x)
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, 0) -> x

The set Q consists of the following terms:

   p(0)
   p(s(x0))
   ack(0, x0)
   ack(s(x0), 0)
   ack(s(x0), s(x1))
   plus(0, x0)
   plus(s(x0), x1)
   plus(x0, s(s(x1)))
   plus(x0, s(0))
   plus(x0, 0)

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

(44) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule PERMUTE(false, s(z0), b) -> PERMUTE(ack(s(z0), s(z0)), p(s(z0)), c) at position [1] we obtained the following new rules [LPAR04]:

   (PERMUTE(false, s(z0), b) -> PERMUTE(ack(s(z0), s(z0)), z0, c),PERMUTE(false, s(z0), b) -> PERMUTE(ack(s(z0), s(z0)), z0, c))


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

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

   PERMUTE(y, x, c) -> PERMUTE(x, y, a)
   PERMUTE(s(x0), y1, a) -> PERMUTE(false, s(x0), b)
   PERMUTE(false, s(z0), b) -> PERMUTE(ack(s(z0), s(z0)), z0, c)

The TRS R consists of the following rules:

   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   ack(s(x), 0) -> ack(x, s(0))
   p(s(x)) -> x
   ack(0, x) -> plus(x, s(0))
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(0)) -> s(x)
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, 0) -> x

The set Q consists of the following terms:

   p(0)
   p(s(x0))
   ack(0, x0)
   ack(s(x0), 0)
   ack(s(x0), s(x1))
   plus(0, x0)
   plus(s(x0), x1)
   plus(x0, s(s(x1)))
   plus(x0, s(0))
   plus(x0, 0)

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

(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.
----------------------------------------

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

   PERMUTE(y, x, c) -> PERMUTE(x, y, a)
   PERMUTE(s(x0), y1, a) -> PERMUTE(false, s(x0), b)
   PERMUTE(false, s(z0), b) -> PERMUTE(ack(s(z0), s(z0)), z0, c)

The TRS R consists of the following rules:

   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   ack(s(x), 0) -> ack(x, s(0))
   ack(0, x) -> plus(x, s(0))
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(0)) -> s(x)
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, 0) -> x

The set Q consists of the following terms:

   p(0)
   p(s(x0))
   ack(0, x0)
   ack(s(x0), 0)
   ack(s(x0), s(x1))
   plus(0, x0)
   plus(s(x0), x1)
   plus(x0, s(s(x1)))
   plus(x0, s(0))
   plus(x0, 0)

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

(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].

   p(0)
   p(s(x0))


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

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

   PERMUTE(y, x, c) -> PERMUTE(x, y, a)
   PERMUTE(s(x0), y1, a) -> PERMUTE(false, s(x0), b)
   PERMUTE(false, s(z0), b) -> PERMUTE(ack(s(z0), s(z0)), z0, c)

The TRS R consists of the following rules:

   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   ack(s(x), 0) -> ack(x, s(0))
   ack(0, x) -> plus(x, s(0))
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(0)) -> s(x)
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, 0) -> x

The set Q consists of the following terms:

   ack(0, x0)
   ack(s(x0), 0)
   ack(s(x0), s(x1))
   plus(0, x0)
   plus(s(x0), x1)
   plus(x0, s(s(x1)))
   plus(x0, s(0))
   plus(x0, 0)

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

(50) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule PERMUTE(false, s(z0), b) -> PERMUTE(ack(s(z0), s(z0)), z0, c) at position [0] we obtained the following new rules [LPAR04]:

   (PERMUTE(false, s(x0), b) -> PERMUTE(ack(x0, ack(s(x0), x0)), x0, c),PERMUTE(false, s(x0), b) -> PERMUTE(ack(x0, ack(s(x0), x0)), x0, c))


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

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

   PERMUTE(y, x, c) -> PERMUTE(x, y, a)
   PERMUTE(s(x0), y1, a) -> PERMUTE(false, s(x0), b)
   PERMUTE(false, s(x0), b) -> PERMUTE(ack(x0, ack(s(x0), x0)), x0, c)

The TRS R consists of the following rules:

   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   ack(s(x), 0) -> ack(x, s(0))
   ack(0, x) -> plus(x, s(0))
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(0)) -> s(x)
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, 0) -> x

The set Q consists of the following terms:

   ack(0, x0)
   ack(s(x0), 0)
   ack(s(x0), s(x1))
   plus(0, x0)
   plus(s(x0), x1)
   plus(x0, s(s(x1)))
   plus(x0, s(0))
   plus(x0, 0)

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

(52) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule PERMUTE(y, x, c) -> PERMUTE(x, y, a) we obtained the following new rules [LPAR04]:

   (PERMUTE(x0, s(y_0), c) -> PERMUTE(s(y_0), x0, a),PERMUTE(x0, s(y_0), c) -> PERMUTE(s(y_0), x0, a))


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

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

   PERMUTE(s(x0), y1, a) -> PERMUTE(false, s(x0), b)
   PERMUTE(false, s(x0), b) -> PERMUTE(ack(x0, ack(s(x0), x0)), x0, c)
   PERMUTE(x0, s(y_0), c) -> PERMUTE(s(y_0), x0, a)

The TRS R consists of the following rules:

   ack(s(x), s(y)) -> ack(x, ack(s(x), y))
   ack(s(x), 0) -> ack(x, s(0))
   ack(0, x) -> plus(x, s(0))
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(x, s(0)) -> s(x)
   plus(x, s(s(y))) -> s(plus(s(x), y))
   plus(x, 0) -> x

The set Q consists of the following terms:

   ack(0, x0)
   ack(s(x0), 0)
   ack(s(x0), s(x1))
   plus(0, x0)
   plus(s(x0), x1)
   plus(x0, s(s(x1)))
   plus(x0, s(0))
   plus(x0, 0)

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

(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:
*PERMUTE(false, s(x0), b) -> PERMUTE(ack(x0, ack(s(x0), x0)), x0, c)
The graph contains the following edges 2 > 2


*PERMUTE(x0, s(y_0), c) -> PERMUTE(s(y_0), x0, a)
The graph contains the following edges 2 >= 1, 1 >= 2


*PERMUTE(s(x0), y1, a) -> PERMUTE(false, s(x0), b)
The graph contains the following edges 1 >= 2


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

(55)
YES