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:
g2(f1(x), y) -> f1(h2(x, y))
h2(x, y) -> g2(x, f1(y))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
g2(f1(x), y) -> f1(h2(x, y))
h2(x, y) -> g2(x, f1(y))
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:
g2(f1(x), y) -> f1(h2(x, y))
h2(x, y) -> g2(x, f1(y))
The set Q consists of the following terms:
g2(f1(x0), x1)
h2(x0, x1)
Q DP problem:
The TRS P consists of the following rules:
H2(x, y) -> G2(x, f1(y))
G2(f1(x), y) -> H2(x, y)
The TRS R consists of the following rules:
g2(f1(x), y) -> f1(h2(x, y))
h2(x, y) -> g2(x, f1(y))
The set Q consists of the following terms:
g2(f1(x0), x1)
h2(x0, x1)
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:
H2(x, y) -> G2(x, f1(y))
G2(f1(x), y) -> H2(x, y)
The TRS R consists of the following rules:
g2(f1(x), y) -> f1(h2(x, y))
h2(x, y) -> g2(x, f1(y))
The set Q consists of the following terms:
g2(f1(x0), x1)
h2(x0, x1)
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.
G2(f1(x), y) -> H2(x, y)
Used argument filtering: H2(x1, x2) = x1
G2(x1, x2) = x1
f1(x1) = f1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
H2(x, y) -> G2(x, f1(y))
The TRS R consists of the following rules:
g2(f1(x), y) -> f1(h2(x, y))
h2(x, y) -> g2(x, f1(y))
The set Q consists of the following terms:
g2(f1(x0), x1)
h2(x0, x1)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.