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:
f3(f3(x, y, z), u, f3(x, y, v)) -> f3(x, y, f3(z, u, v))
f3(x, y, y) -> y
f3(x, y, g1(y)) -> x
f3(x, x, y) -> x
f3(g1(x), x, y) -> y
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f3(f3(x, y, z), u, f3(x, y, v)) -> f3(x, y, f3(z, u, v))
f3(x, y, y) -> y
f3(x, y, g1(y)) -> x
f3(x, x, y) -> x
f3(g1(x), x, y) -> y
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F3(f3(x, y, z), u, f3(x, y, v)) -> F3(x, y, f3(z, u, v))
F3(f3(x, y, z), u, f3(x, y, v)) -> F3(z, u, v)
The TRS R consists of the following rules:
f3(f3(x, y, z), u, f3(x, y, v)) -> f3(x, y, f3(z, u, v))
f3(x, y, y) -> y
f3(x, y, g1(y)) -> x
f3(x, x, y) -> x
f3(g1(x), x, y) -> y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
F3(f3(x, y, z), u, f3(x, y, v)) -> F3(x, y, f3(z, u, v))
F3(f3(x, y, z), u, f3(x, y, v)) -> F3(z, u, v)
The TRS R consists of the following rules:
f3(f3(x, y, z), u, f3(x, y, v)) -> f3(x, y, f3(z, u, v))
f3(x, y, y) -> y
f3(x, y, g1(y)) -> x
f3(x, x, y) -> x
f3(g1(x), x, y) -> y
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.
F3(f3(x, y, z), u, f3(x, y, v)) -> F3(x, y, f3(z, u, v))
F3(f3(x, y, z), u, f3(x, y, v)) -> F3(z, u, v)
Used argument filtering: F3(x1, x2, x3) = x1
f3(x1, x2, x3) = f2(x1, x3)
g1(x1) = g
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f3(f3(x, y, z), u, f3(x, y, v)) -> f3(x, y, f3(z, u, v))
f3(x, y, y) -> y
f3(x, y, g1(y)) -> x
f3(x, x, y) -> x
f3(g1(x), x, y) -> y
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.