Termination w.r.t. Q of the given QTRS could not be shown:

(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 transformation)

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 transformation)

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) 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.

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
ACK(s(x), 0) → ACK(x, s(0))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACK(x1, x2)  =  ACK(x1)
s(x1)  =  s(x1)
ack(x1, x2)  =  ack
0  =  0
plus(x1, x2)  =  x1

Lexicographic path order with status [LPO].
Precedence:

ack > 0 > ACK₁ > s₁

Status:

ack: []
ACK₁: [1]
s₁: [1]
0: []

The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

Q DP problem:
The TRS P consists of the following rules:

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.

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACK(s(x), s(y)) → ACK(s(x), y)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

s₁ > ACK₂

Status:

ACK₂: [1,2]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(10) Obligation:

Q DP problem:
P is empty.
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.

(11) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(12) TRUE

(13) 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.