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:
a1(b1(x)) -> a1(c1(b1(x)))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
a1(b1(x)) -> a1(c1(b1(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:
a1(b1(x)) -> a1(c1(b1(x)))
The set Q consists of the following terms:
a1(b1(x0))
Q DP problem:
The TRS P consists of the following rules:
A1(b1(x)) -> A1(c1(b1(x)))
The TRS R consists of the following rules:
a1(b1(x)) -> a1(c1(b1(x)))
The set Q consists of the following terms:
a1(b1(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:
A1(b1(x)) -> A1(c1(b1(x)))
The TRS R consists of the following rules:
a1(b1(x)) -> a1(c1(b1(x)))
The set Q consists of the following terms:
a1(b1(x0))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.