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:
g1(a) -> g1(b)
b -> f2(a, a)
f2(a, a) -> g1(d)
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
g1(a) -> g1(b)
b -> f2(a, a)
f2(a, a) -> g1(d)
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:
g1(a) -> g1(b)
b -> f2(a, a)
f2(a, a) -> g1(d)
The set Q consists of the following terms:
g1(a)
b
f2(a, a)
Q DP problem:
The TRS P consists of the following rules:
F2(a, a) -> G1(d)
B -> F2(a, a)
G1(a) -> B
G1(a) -> G1(b)
The TRS R consists of the following rules:
g1(a) -> g1(b)
b -> f2(a, a)
f2(a, a) -> g1(d)
The set Q consists of the following terms:
g1(a)
b
f2(a, a)
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(a, a) -> G1(d)
B -> F2(a, a)
G1(a) -> B
G1(a) -> G1(b)
The TRS R consists of the following rules:
g1(a) -> g1(b)
b -> f2(a, a)
f2(a, a) -> g1(d)
The set Q consists of the following terms:
g1(a)
b
f2(a, a)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 4 less nodes.