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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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 2 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

POL(*2(x1, x2)) = 1 + 2·x1 + x2   
POL(*12(x1, x2)) = x1   

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

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.