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(i1(x), x) -> 1
*2(1, y) -> y
*2(x, 0) -> 0
*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(i1(x), x) -> 1
*2(1, y) -> y
*2(x, 0) -> 0
*2(*2(x, y), z) -> *2(x, *2(y, z))

Q is empty.

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(i1(x), x) -> 1
*2(1, y) -> y
*2(x, 0) -> 0
*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
      ↳ QDPAfsSolverProof

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(i1(x), x) -> 1
*2(1, y) -> y
*2(x, 0) -> 0
*2(*2(x, y), z) -> *2(x, *2(y, z))

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(*2(x, y), z) -> *12(y, z)
*12(*2(x, y), z) -> *12(x, *2(y, z))
Used argument filtering: *12(x1, x2)  =  x1
*2(x1, x2)  =  *2(x1, x2)
i1(x1)  =  i
1  =  1
0  =  0
Used ordering: Quasi Precedence: *_2 > 1 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ PisEmptyProof

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

*2(i1(x), x) -> 1
*2(1, y) -> y
*2(x, 0) -> 0
*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.