Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.


↳ QTRS
  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

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

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.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

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

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.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof

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

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

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


+1(x, +(y, z)) → +1(x, y)
+1(x, minus(y)) → +1(minus(x), y)
+1(x, +(y, z)) → +1(+(x, y), z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
+1(x1, x2)  =  +1(x1, x2)
+(x1, x2)  =  +(x1, x2)
minus(x1)  =  minus(x1)
1  =  1
0  =  0

Lexicographic path order with status [19].
Quasi-Precedence:
[+^12, +2] > [minus1, 1]
[+^12, +2] > 0

Status:
minus1: [1]
+2: [2,1]
1: multiset
0: multiset
+^12: [2,1]


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ PisEmptyProof

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