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:
f2(s1(X), X) -> f2(X, a1(X))
f2(X, c1(X)) -> f2(s1(X), X)
f2(X, X) -> c1(X)
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(s1(X), X) -> f2(X, a1(X))
f2(X, c1(X)) -> f2(s1(X), X)
f2(X, X) -> c1(X)
Q is empty.
The TRS is non-overlapping. Hence, we can switch to innermost.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(s1(X), X) -> f2(X, a1(X))
f2(X, c1(X)) -> f2(s1(X), X)
f2(X, X) -> c1(X)
The set Q consists of the following terms:
f2(s1(x0), x0)
f2(x0, c1(x0))
f2(x0, x0)
Q DP problem:
The TRS P consists of the following rules:
F2(X, c1(X)) -> F2(s1(X), X)
F2(s1(X), X) -> F2(X, a1(X))
The TRS R consists of the following rules:
f2(s1(X), X) -> f2(X, a1(X))
f2(X, c1(X)) -> f2(s1(X), X)
f2(X, X) -> c1(X)
The set Q consists of the following terms:
f2(s1(x0), x0)
f2(x0, c1(x0))
f2(x0, x0)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F2(X, c1(X)) -> F2(s1(X), X)
F2(s1(X), X) -> F2(X, a1(X))
The TRS R consists of the following rules:
f2(s1(X), X) -> f2(X, a1(X))
f2(X, c1(X)) -> f2(s1(X), X)
f2(X, X) -> c1(X)
The set Q consists of the following terms:
f2(s1(x0), x0)
f2(x0, c1(x0))
f2(x0, x0)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 2 less nodes.