Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

*2(X, +2(Y, 1)) -> +2(*2(X, +2(Y, *2(1, 0))), X)
*2(X, 1) -> X
*2(X, 0) -> X
*2(X, 0) -> 0

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

*2(X, +2(Y, 1)) -> +2(*2(X, +2(Y, *2(1, 0))), X)
*2(X, 1) -> X
*2(X, 0) -> X
*2(X, 0) -> 0

Q is empty.

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

*12(X, +2(Y, 1)) -> *12(1, 0)
*12(X, +2(Y, 1)) -> *12(X, +2(Y, *2(1, 0)))

The TRS R consists of the following rules:

*2(X, +2(Y, 1)) -> +2(*2(X, +2(Y, *2(1, 0))), X)
*2(X, 1) -> X
*2(X, 0) -> X
*2(X, 0) -> 0

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(X, +2(Y, 1)) -> *12(1, 0)
*12(X, +2(Y, 1)) -> *12(X, +2(Y, *2(1, 0)))

The TRS R consists of the following rules:

*2(X, +2(Y, 1)) -> +2(*2(X, +2(Y, *2(1, 0))), X)
*2(X, 1) -> X
*2(X, 0) -> X
*2(X, 0) -> 0

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP

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

*12(X, +2(Y, 1)) -> *12(X, +2(Y, *2(1, 0)))

The TRS R consists of the following rules:

*2(X, +2(Y, 1)) -> +2(*2(X, +2(Y, *2(1, 0))), X)
*2(X, 1) -> X
*2(X, 0) -> X
*2(X, 0) -> 0

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