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

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

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

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

*2(x0, +2(x1, x2))


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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

*2(x0, +2(x1, x2))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

*2(x0, +2(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


*12(x, +2(y, z)) -> *12(x, z)
*12(x, +2(y, z)) -> *12(x, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
*12(x1, x2)  =  *11(x2)
+2(x1, x2)  =  +2(x1, x2)

Lexicographic Path Order [19].
Precedence:
+2 > *^11

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

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

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

The set Q consists of the following terms:

*2(x0, +2(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.