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)) -> b1(b1(a1(x)))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
a1(b1(x)) -> b1(b1(a1(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)) -> b1(b1(a1(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(x)
The TRS R consists of the following rules:
a1(b1(x)) -> b1(b1(a1(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
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
A1(b1(x)) -> A1(x)
The TRS R consists of the following rules:
a1(b1(x)) -> b1(b1(a1(x)))
The set Q consists of the following terms:
a1(b1(x0))
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
A1(b1(x)) -> A1(x)
Used argument filtering: A1(x1) = x1
b1(x1) = b1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
a1(b1(x)) -> b1(b1(a1(x)))
The set Q consists of the following terms:
a1(b1(x0))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.