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:
g1(h1(g1(x))) -> g1(x)
g1(g1(x)) -> g1(h1(g1(x)))
h1(h1(x)) -> h1(f2(h1(x), x))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g1(h1(g1(x))) -> g1(x)
g1(g1(x)) -> g1(h1(g1(x)))
h1(h1(x)) -> h1(f2(h1(x), x))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
G1(g1(x)) -> H1(g1(x))
G1(g1(x)) -> G1(h1(g1(x)))
H1(h1(x)) -> H1(f2(h1(x), x))
The TRS R consists of the following rules:
g1(h1(g1(x))) -> g1(x)
g1(g1(x)) -> g1(h1(g1(x)))
h1(h1(x)) -> h1(f2(h1(x), 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:
G1(g1(x)) -> H1(g1(x))
G1(g1(x)) -> G1(h1(g1(x)))
H1(h1(x)) -> H1(f2(h1(x), x))
The TRS R consists of the following rules:
g1(h1(g1(x))) -> g1(x)
g1(g1(x)) -> g1(h1(g1(x)))
h1(h1(x)) -> h1(f2(h1(x), 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:
G1(g1(x)) -> G1(h1(g1(x)))
The TRS R consists of the following rules:
g1(h1(g1(x))) -> g1(x)
g1(g1(x)) -> g1(h1(g1(x)))
h1(h1(x)) -> h1(f2(h1(x), 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.
G1(g1(x)) -> G1(h1(g1(x)))
Used argument filtering: G1(x1) = x1
g1(x1) = g
h1(x1) = h
Used ordering: Precedence:
g > h
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
g1(h1(g1(x))) -> g1(x)
g1(g1(x)) -> g1(h1(g1(x)))
h1(h1(x)) -> h1(f2(h1(x), 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.