Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
f3(0, 1, x) -> f3(g1(x), g1(x), x)
f3(g1(x), y, z) -> g1(f3(x, y, z))
f3(x, g1(y), z) -> g1(f3(x, y, z))
f3(x, y, g1(z)) -> g1(f3(x, y, z))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f3(0, 1, x) -> f3(g1(x), g1(x), x)
f3(g1(x), y, z) -> g1(f3(x, y, z))
f3(x, g1(y), z) -> g1(f3(x, y, z))
f3(x, y, g1(z)) -> g1(f3(x, y, z))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F3(g1(x), y, z) -> F3(x, y, z)
F3(0, 1, x) -> F3(g1(x), g1(x), x)
F3(x, y, g1(z)) -> F3(x, y, z)
F3(x, g1(y), z) -> F3(x, y, z)
The TRS R consists of the following rules:
f3(0, 1, x) -> f3(g1(x), g1(x), x)
f3(g1(x), y, z) -> g1(f3(x, y, z))
f3(x, g1(y), z) -> g1(f3(x, y, z))
f3(x, y, g1(z)) -> g1(f3(x, y, z))
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(g1(x), y, z) -> F3(x, y, z)
F3(0, 1, x) -> F3(g1(x), g1(x), x)
F3(x, y, g1(z)) -> F3(x, y, z)
F3(x, g1(y), z) -> F3(x, y, z)
The TRS R consists of the following rules:
f3(0, 1, x) -> f3(g1(x), g1(x), x)
f3(g1(x), y, z) -> g1(f3(x, y, z))
f3(x, g1(y), z) -> g1(f3(x, y, z))
f3(x, y, g1(z)) -> g1(f3(x, y, z))
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(x, y, g1(z)) -> F3(x, y, z)
Used argument filtering: F3(x1, x2, x3) = x3
g1(x1) = g1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F3(g1(x), y, z) -> F3(x, y, z)
F3(0, 1, x) -> F3(g1(x), g1(x), x)
F3(x, g1(y), z) -> F3(x, y, z)
The TRS R consists of the following rules:
f3(0, 1, x) -> f3(g1(x), g1(x), x)
f3(g1(x), y, z) -> g1(f3(x, y, z))
f3(x, g1(y), z) -> g1(f3(x, y, z))
f3(x, y, g1(z)) -> g1(f3(x, y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.