Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(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(+(x, y)) → +¹(minus(y), minus(x))
MINUS(+(x, y)) → MINUS(y)
MINUS(+(x, y)) → MINUS(x)
*¹(x, +(y, z)) → +¹(*(x, y), *(x, z))
*¹(x, +(y, z)) → *¹(x, y)
*¹(x, +(y, z)) → *¹(x, z)
*¹(x, minus(y)) → MINUS(*(x, y))
*¹(x, minus(y)) → *¹(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(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 2 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

MINUS(+(x, y)) → MINUS(x)
MINUS(+(x, y)) → MINUS(y)

The TRS R consists of the following rules:

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, y))

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.

MINUS(+(x, y)) → MINUS(x)
MINUS(+(x, y)) → MINUS(y)

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

[MINUS₁, +₂]

Status:

MINUS₁: multiset
+₂: multiset

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, y))

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

*¹(x, +(y, z)) → *¹(x, z)
*¹(x, +(y, z)) → *¹(x, y)
*¹(x, minus(y)) → *¹(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, y))

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

*¹(x, +(y, z)) → *¹(x, z)
*¹(x, +(y, z)) → *¹(x, y)
*¹(x, minus(y)) → *¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
*¹(x1, x2)  =  x2
+(x1, x2)  =  +(x1, x2)
minus(x1)  =  minus(x1)

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

trivial

Status:

minus₁: multiset
+₂: multiset

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, y))

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE