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:
app2(f, a) -> app2(f, b)
app2(g, b) -> app2(g, a)
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(f, a) -> app2(f, b)
app2(g, b) -> app2(g, a)
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:
app2(f, a) -> app2(f, b)
app2(g, b) -> app2(g, a)
The set Q consists of the following terms:
app2(f, a)
app2(g, b)
Q DP problem:
The TRS P consists of the following rules:
APP2(g, b) -> APP2(g, a)
APP2(f, a) -> APP2(f, b)
The TRS R consists of the following rules:
app2(f, a) -> app2(f, b)
app2(g, b) -> app2(g, a)
The set Q consists of the following terms:
app2(f, a)
app2(g, b)
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:
APP2(g, b) -> APP2(g, a)
APP2(f, a) -> APP2(f, b)
The TRS R consists of the following rules:
app2(f, a) -> app2(f, b)
app2(g, b) -> app2(g, a)
The set Q consists of the following terms:
app2(f, a)
app2(g, b)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 2 less nodes.