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:
f(X, g(X)) → f(1, g(X))
g(1) → g(0)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(X, g(X)) → f(1, g(X))
g(1) → g(0)
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:
F(X, g(X)) → F(1, g(X))
G(1) → G(0)
The TRS R consists of the following rules:
f(X, g(X)) → f(1, g(X))
g(1) → g(0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
F(X, g(X)) → F(1, g(X))
G(1) → G(0)
The TRS R consists of the following rules:
f(X, g(X)) → f(1, g(X))
g(1) → g(0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(X, g(X)) → F(1, g(X))
G(1) → G(0)
The TRS R consists of the following rules:
f(X, g(X)) → f(1, g(X))
g(1) → g(0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(X, g(X)) → F(1, g(X))
The TRS R consists of the following rules:
f(X, g(X)) → f(1, g(X))
g(1) → g(0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.