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 QDPOrderProof (⇔)
    ↳7 QDP
  ↳8 QDP
    ↳9 QDPOrderProof (⇔)
    ↳10 QDP
    ↳11 PisEmptyProof (⇔)
    ↳12 TRUE
  ↳13 QDP
    ↳14 QDPOrderProof (⇔)
    ↳15 QDP
    ↳16 PisEmptyProof (⇔)
    ↳17 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0)))

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(s(x), s(y)) → MINUS(x, y)
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) → MINUS(x, y)
PLUS(s(x), y) → PLUS(x, y)
PLUS(minus(x, s(0)), minus(y, s(s(z)))) → PLUS(minus(y, s(s(z))), minus(x, s(0)))
PLUS(plus(x, s(0)), plus(y, s(s(z)))) → PLUS(plus(y, s(s(z))), plus(x, s(0)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0)))

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 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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

PLUS(minus(x, s(0)), minus(y, s(s(z)))) → PLUS(minus(y, s(s(z))), minus(x, s(0)))
PLUS(s(x), y) → PLUS(x, y)
PLUS(plus(x, s(0)), plus(y, s(s(z)))) → PLUS(plus(y, s(s(z))), plus(x, s(0)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0)))

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
PLUS2 > plus2 > s1
PLUS2 > plus2 > 0
quot2 > s1
quot2 > 0

The following usable rules [FROCOS05] were oriented:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0)))

(7) Obligation:

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

PLUS(minus(x, s(0)), minus(y, s(s(z)))) → PLUS(minus(y, s(s(z))), minus(x, s(0)))
PLUS(plus(x, s(0)), plus(y, s(s(z)))) → PLUS(plus(y, s(s(z))), plus(x, s(0)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0)))

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

(8) 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(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0)))

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.


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

Recursive Path Order [RPO].
Precedence:
quot1 > s1 > MINUS1 > minus1
quot1 > 0 > minus1
plus2 > s1 > MINUS1 > minus1
plus2 > 0 > minus1

The following usable rules [FROCOS05] were oriented:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0)))

(10) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0)))

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:

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0)))

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.


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

Recursive Path Order [RPO].
Precedence:
quot2 > s1 > minus1 > 0
plus2 > s1 > minus1 > 0

The following usable rules [FROCOS05] were oriented:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0)))

(15) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) → plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) → plus(plus(y, s(s(z))), plus(x, s(0)))

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

(16) PisEmptyProof (EQUIVALENT transformation)

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

(17) TRUE