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:

+2(-2(x, y), z) -> -2(+2(x, z), y)
-2(+2(x, y), y) -> x

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

+2(-2(x, y), z) -> -2(+2(x, z), y)
-2(+2(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:

+12(-2(x, y), z) -> +12(x, z)
+12(-2(x, y), z) -> -12(+2(x, z), y)

The TRS R consists of the following rules:

+2(-2(x, y), z) -> -2(+2(x, z), y)
-2(+2(x, y), y) -> x

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

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

+12(-2(x, y), z) -> +12(x, z)
+12(-2(x, y), z) -> -12(+2(x, z), y)

The TRS R consists of the following rules:

+2(-2(x, y), z) -> -2(+2(x, z), y)
-2(+2(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
      ↳ DependencyGraphProof
QDP
          ↳ QDPOrderProof

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

+12(-2(x, y), z) -> +12(x, z)

The TRS R consists of the following rules:

+2(-2(x, y), z) -> -2(+2(x, z), y)
-2(+2(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.


+12(-2(x, y), z) -> +12(x, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( +12(x1, x2) ) = max{0, x1 - 1}


POL( -2(x1, x2) ) = x1 + 2



The following usable rules [14] were oriented: none



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

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

+2(-2(x, y), z) -> -2(+2(x, z), y)
-2(+2(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.