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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
  ↳6 QDP
    ↳7 QDPOrderProof (⇔)
    ↳8 QDP
    ↳9 QDPOrderProof (⇔)
    ↳10 QDP
    ↳11 PisEmptyProof (⇔)
    ↳12 TRUE
  ↳13 QDP
    ↳14 QDPOrderProof (⇔)
    ↳15 QDP
    ↳16 QDPOrderProof (⇔)
    ↳17 QDP
    ↳18 PisEmptyProof (⇔)
    ↳19 TRUE
  ↳20 QDP

(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)  =  x1
0  =  0
plus(x1, x2)  =  plus(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:
[ACK1, 0] > plus2 > s1

Status:
plus2: [2,1]
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:

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: Lexicographic path order with status [LPO].
Quasi-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:

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

Lexicographic path order with status [LPO].
Quasi-Precedence:
s1 > MINUS1
s1 > plus

Status:
plus: []
s1: [1]
MINUS1: [1]
0: []


The following usable rules [FROCOS05] were oriented: none

(15) Obligation:

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

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.

(16) 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)  =  MINUS(x1, x2)
minus(x1, x2)  =  minus(x1)
plus(x1, x2)  =  x2
s(x1)  =  s
0  =  0

Lexicographic path order with status [LPO].
Quasi-Precedence:
minus1 > [MINUS2, s]
0 > [MINUS2, s]

Status:
minus1: [1]
MINUS2: [2,1]
s: []
0: []


The following usable rules [FROCOS05] were oriented:

plus(s(x), y) → s(plus(y, x))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))

(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.