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(f1(x)) -> f1(g2(f1(x), x))
f1(f1(x)) -> f1(h2(f1(x), f1(x)))
g2(x, y) -> y
h2(x, x) -> g2(x, 0)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(f1(x)) -> f1(g2(f1(x), x))
f1(f1(x)) -> f1(h2(f1(x), f1(x)))
g2(x, y) -> y
h2(x, x) -> g2(x, 0)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F1(f1(x)) -> F1(g2(f1(x), x))
F1(f1(x)) -> H2(f1(x), f1(x))
H2(x, x) -> G2(x, 0)
F1(f1(x)) -> G2(f1(x), x)
F1(f1(x)) -> F1(h2(f1(x), f1(x)))
The TRS R consists of the following rules:
f1(f1(x)) -> f1(g2(f1(x), x))
f1(f1(x)) -> f1(h2(f1(x), f1(x)))
g2(x, y) -> y
h2(x, x) -> g2(x, 0)
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(f1(x)) -> F1(g2(f1(x), x))
F1(f1(x)) -> H2(f1(x), f1(x))
H2(x, x) -> G2(x, 0)
F1(f1(x)) -> G2(f1(x), x)
F1(f1(x)) -> F1(h2(f1(x), f1(x)))
The TRS R consists of the following rules:
f1(f1(x)) -> f1(g2(f1(x), x))
f1(f1(x)) -> f1(h2(f1(x), f1(x)))
g2(x, y) -> y
h2(x, x) -> g2(x, 0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
F1(f1(x)) -> F1(g2(f1(x), x))
F1(f1(x)) -> F1(h2(f1(x), f1(x)))
The TRS R consists of the following rules:
f1(f1(x)) -> f1(g2(f1(x), x))
f1(f1(x)) -> f1(h2(f1(x), f1(x)))
g2(x, y) -> y
h2(x, x) -> g2(x, 0)
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(f1(x)) -> F1(g2(f1(x), x))
F1(f1(x)) -> F1(h2(f1(x), f1(x)))
Used argument filtering: F1(x1) = x1
f1(x1) = f1(x1)
g2(x1, x2) = x2
h2(x1, x2) = h
0 = 0
Used ordering: Quasi Precedence:
f_1 > h > 0
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f1(f1(x)) -> f1(g2(f1(x), x))
f1(f1(x)) -> f1(h2(f1(x), f1(x)))
g2(x, y) -> y
h2(x, x) -> g2(x, 0)
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.