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(minus1(y), y)) -> *2(minus1(*2(y, y)), x)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

*2(x, *2(minus1(y), y)) -> *2(minus1(*2(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(x, *2(minus1(y), y)) -> *12(minus1(*2(y, y)), x)
*12(x, *2(minus1(y), y)) -> *12(y, y)

The TRS R consists of the following rules:

*2(x, *2(minus1(y), y)) -> *2(minus1(*2(y, y)), x)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

*12(x, *2(minus1(y), y)) -> *12(minus1(*2(y, y)), x)
*12(x, *2(minus1(y), y)) -> *12(y, y)

The TRS R consists of the following rules:

*2(x, *2(minus1(y), y)) -> *2(minus1(*2(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(x, *2(minus1(y), y)) -> *12(minus1(*2(y, y)), x)
*12(x, *2(minus1(y), y)) -> *12(y, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( minus1(x1) ) = max{0, -1}


POL( *12(x1, x2) ) = 3x1 + 3x2 + 1


POL( *2(x1, x2) ) = 3x2 + 2



The following usable rules [14] were oriented: none



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

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

*2(x, *2(minus1(y), y)) -> *2(minus1(*2(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.