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:

+(-(x, y), z) → -(+(x, z), y)
-(+(x, y), y) → x

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

+(-(x, y), z) → -(+(x, z), y)
-(+(x, y), y) → 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 1 less node.

↳ 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, z)

The TRS R consists of the following rules:

+(-(x, y), z) → -(+(x, z), y)
-(+(x, y), y) → 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, 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)
-(x1, x2)  =  -(x1, x2)

Recursive Path Order [2].
Precedence:
-2 > +^11

The following usable rules [14] were oriented: none



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

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

+(-(x, y), z) → -(+(x, z), y)
-(+(x, y), y) → 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.