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:
f2(a, x) -> f2(b, f2(c, x))
f2(a, f2(b, x)) -> f2(b, f2(a, x))
f2(d, f2(c, x)) -> f2(d, f2(a, x))
f2(a, f2(c, x)) -> f2(c, f2(a, x))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(a, x) -> f2(b, f2(c, x))
f2(a, f2(b, x)) -> f2(b, f2(a, x))
f2(d, f2(c, x)) -> f2(d, f2(a, x))
f2(a, f2(c, x)) -> f2(c, f2(a, x))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F2(a, f2(c, x)) -> F2(c, f2(a, x))
F2(a, x) -> F2(b, f2(c, x))
F2(d, f2(c, x)) -> F2(d, f2(a, x))
F2(a, x) -> F2(c, x)
F2(a, f2(b, x)) -> F2(a, x)
F2(a, f2(b, x)) -> F2(b, f2(a, x))
F2(a, f2(c, x)) -> F2(a, x)
F2(d, f2(c, x)) -> F2(a, x)
The TRS R consists of the following rules:
f2(a, x) -> f2(b, f2(c, x))
f2(a, f2(b, x)) -> f2(b, f2(a, x))
f2(d, f2(c, x)) -> f2(d, f2(a, x))
f2(a, f2(c, x)) -> f2(c, f2(a, 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:
F2(a, f2(c, x)) -> F2(c, f2(a, x))
F2(a, x) -> F2(b, f2(c, x))
F2(d, f2(c, x)) -> F2(d, f2(a, x))
F2(a, x) -> F2(c, x)
F2(a, f2(b, x)) -> F2(a, x)
F2(a, f2(b, x)) -> F2(b, f2(a, x))
F2(a, f2(c, x)) -> F2(a, x)
F2(d, f2(c, x)) -> F2(a, x)
The TRS R consists of the following rules:
f2(a, x) -> f2(b, f2(c, x))
f2(a, f2(b, x)) -> f2(b, f2(a, x))
f2(d, f2(c, x)) -> f2(d, f2(a, x))
f2(a, f2(c, x)) -> f2(c, f2(a, x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 5 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F2(a, f2(b, x)) -> F2(a, x)
F2(a, f2(c, x)) -> F2(a, x)
The TRS R consists of the following rules:
f2(a, x) -> f2(b, f2(c, x))
f2(a, f2(b, x)) -> f2(b, f2(a, x))
f2(d, f2(c, x)) -> f2(d, f2(a, x))
f2(a, f2(c, x)) -> f2(c, f2(a, 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.
F2(a, f2(b, x)) -> F2(a, x)
F2(a, f2(c, x)) -> F2(a, x)
Used argument filtering: F2(x1, x2) = x2
f2(x1, x2) = f1(x2)
b = b
c = c
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f2(a, x) -> f2(b, f2(c, x))
f2(a, f2(b, x)) -> f2(b, f2(a, x))
f2(d, f2(c, x)) -> f2(d, f2(a, x))
f2(a, f2(c, x)) -> f2(c, f2(a, 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.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F2(d, f2(c, x)) -> F2(d, f2(a, x))
The TRS R consists of the following rules:
f2(a, x) -> f2(b, f2(c, x))
f2(a, f2(b, x)) -> f2(b, f2(a, x))
f2(d, f2(c, x)) -> f2(d, f2(a, x))
f2(a, f2(c, x)) -> f2(c, f2(a, x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.