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.