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:

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(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))

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:

MINUS(minus(x, y), z) → MINUS(x, plus(y, z))
MINUS(minus(x, y), z) → PLUS(y, z)
MINUS(s(x), s(y)) → MINUS(x, y)
PLUS(s(x), y) → PLUS(x, s(y))
PLUS(s(x), y) → PLUS(y, x)
DIV(x, y) → QUOT(x, y, 0)
QUOT(s(x), s(y), z) → QUOT(minus(p(ack(0, x)), y), s(y), s(z))
QUOT(s(x), s(y), z) → MINUS(p(ack(0, x)), y)
QUOT(s(x), s(y), z) → P(ack(0, x))
QUOT(s(x), s(y), z) → ACK(0, 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))
ACK(s(x), s(y)) → ACK(s(x), y)

The TRS R consists of the following rules:

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(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))

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 4 SCCs with 6 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

PLUS(s(x), y) → PLUS(y, x)
PLUS(s(x), y) → PLUS(x, s(y))

The TRS R consists of the following rules:

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(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))

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:

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(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))

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)  =  plus(x2)

Recursive path order with status [RPO].
Quasi-Precedence:

[ACK₁, s₁, ack, 0, plus₁]

Status:

ack: multiset
plus₁: [1]
ACK₁: multiset
s₁: multiset
0: multiset

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:

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(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))

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: Recursive path order with status [RPO].
Quasi-Precedence:

s₁ > ACK₂

Status:

ACK₂: [1,2]
s₁: multiset

The following usable rules [FROCOS05] were oriented: none

(10) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(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))

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:

MINUS(s(x), s(y)) → MINUS(x, y)
MINUS(minus(x, y), z) → MINUS(x, plus(y, z))

The TRS R consists of the following rules:

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(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))

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

(14) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MINUS(minus(x, y), z) → MINUS(x, plus(y, z))

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

Recursive path order with status [RPO].
Quasi-Precedence:

minus₂ > plus₁

Status:

minus₂: multiset
plus₁: multiset
0: multiset

The following usable rules [FROCOS05] were oriented: none

(15) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(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))

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MINUS(s(x), s(y)) → MINUS(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Quasi-Precedence:

[MINUS₂, s₁]

Status:

MINUS₂: [1,2]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(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))

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) Obligation:

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

QUOT(s(x), s(y), z) → QUOT(minus(p(ack(0, x)), y), s(y), s(z))

The TRS R consists of the following rules:

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(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))

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