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:
f3(0, 1, X) -> f3(g2(X, X), X, X)
g2(X, Y) -> X
g2(X, Y) -> Y
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f3(0, 1, X) -> f3(g2(X, X), X, X)
g2(X, Y) -> X
g2(X, Y) -> Y
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F3(0, 1, X) -> F3(g2(X, X), X, X)
F3(0, 1, X) -> G2(X, X)
The TRS R consists of the following rules:
f3(0, 1, X) -> f3(g2(X, X), X, X)
g2(X, Y) -> X
g2(X, Y) -> Y
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:
F3(0, 1, X) -> F3(g2(X, X), X, X)
F3(0, 1, X) -> G2(X, X)
The TRS R consists of the following rules:
f3(0, 1, X) -> f3(g2(X, X), X, X)
g2(X, Y) -> X
g2(X, Y) -> Y
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:
F3(0, 1, X) -> F3(g2(X, X), X, X)
The TRS R consists of the following rules:
f3(0, 1, X) -> f3(g2(X, X), X, X)
g2(X, Y) -> X
g2(X, Y) -> Y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.