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:
f1(a) -> b
f1(c) -> d
f1(g2(x, y)) -> g2(f1(x), f1(y))
f1(h2(x, y)) -> g2(h2(y, f1(x)), h2(x, f1(y)))
g2(x, x) -> h2(e, x)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(a) -> b
f1(c) -> d
f1(g2(x, y)) -> g2(f1(x), f1(y))
f1(h2(x, y)) -> g2(h2(y, f1(x)), h2(x, f1(y)))
g2(x, x) -> h2(e, x)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F1(h2(x, y)) -> F1(y)
F1(g2(x, y)) -> F1(y)
F1(g2(x, y)) -> G2(f1(x), f1(y))
F1(h2(x, y)) -> G2(h2(y, f1(x)), h2(x, f1(y)))
F1(g2(x, y)) -> F1(x)
F1(h2(x, y)) -> F1(x)
The TRS R consists of the following rules:
f1(a) -> b
f1(c) -> d
f1(g2(x, y)) -> g2(f1(x), f1(y))
f1(h2(x, y)) -> g2(h2(y, f1(x)), h2(x, f1(y)))
g2(x, x) -> h2(e, x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F1(h2(x, y)) -> F1(y)
F1(g2(x, y)) -> F1(y)
F1(g2(x, y)) -> G2(f1(x), f1(y))
F1(h2(x, y)) -> G2(h2(y, f1(x)), h2(x, f1(y)))
F1(g2(x, y)) -> F1(x)
F1(h2(x, y)) -> F1(x)
The TRS R consists of the following rules:
f1(a) -> b
f1(c) -> d
f1(g2(x, y)) -> g2(f1(x), f1(y))
f1(h2(x, y)) -> g2(h2(y, f1(x)), h2(x, f1(y)))
g2(x, x) -> h2(e, x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
F1(h2(x, y)) -> F1(y)
F1(g2(x, y)) -> F1(y)
F1(g2(x, y)) -> F1(x)
F1(h2(x, y)) -> F1(x)
The TRS R consists of the following rules:
f1(a) -> b
f1(c) -> d
f1(g2(x, y)) -> g2(f1(x), f1(y))
f1(h2(x, y)) -> g2(h2(y, f1(x)), h2(x, f1(y)))
g2(x, x) -> h2(e, x)
Q is empty.
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.
F1(h2(x, y)) -> F1(y)
F1(g2(x, y)) -> F1(y)
F1(g2(x, y)) -> F1(x)
F1(h2(x, y)) -> F1(x)
Used argument filtering: F1(x1) = x1
h2(x1, x2) = h2(x1, x2)
g2(x1, x2) = g2(x1, x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f1(a) -> b
f1(c) -> d
f1(g2(x, y)) -> g2(f1(x), f1(y))
f1(h2(x, y)) -> g2(h2(y, f1(x)), h2(x, f1(y)))
g2(x, x) -> h2(e, x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.