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:

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

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:

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

The TRS R consists of the following rules:

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

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 1 SCC with 3 less nodes.

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(5) 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, y), z)
+¹(x, +(y, z)) → +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)  =  x1
0  =  0
1  =  1

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(6) Obligation:

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

+¹(x, minus(y)) → +¹(minus(x), y)

The TRS R consists of the following rules:

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

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.

+¹(x, minus(y)) → +¹(minus(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
minus(x1)  =  minus(x1)
0  =  0

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

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

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

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

(9) PisEmptyProof (EQUIVALENT transformation)

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

(10) TRUE