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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

*12(x, *2(y, z)) -> *12(x, y)
*12(x, *2(y, z)) -> *12(*2(x, y), z)
Used argument filtering: *12(x1, x2)  =  x2
*2(x1, x2)  =  *2(x1, x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ 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), z)
*2(x, x) -> 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.