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, h1(x)) -> f2(g1(x), h1(x))
h1(g1(x)) -> h1(a)
g1(h1(x)) -> g1(x)
h1(h1(x)) -> x
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(a, h1(x)) -> f2(g1(x), h1(x))
h1(g1(x)) -> h1(a)
g1(h1(x)) -> g1(x)
h1(h1(x)) -> x
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
G1(h1(x)) -> G1(x)
F2(a, h1(x)) -> G1(x)
H1(g1(x)) -> H1(a)
F2(a, h1(x)) -> F2(g1(x), h1(x))
The TRS R consists of the following rules:
f2(a, h1(x)) -> f2(g1(x), h1(x))
h1(g1(x)) -> h1(a)
g1(h1(x)) -> g1(x)
h1(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(h1(x)) -> G1(x)
F2(a, h1(x)) -> G1(x)
H1(g1(x)) -> H1(a)
F2(a, h1(x)) -> F2(g1(x), h1(x))
The TRS R consists of the following rules:
f2(a, h1(x)) -> f2(g1(x), h1(x))
h1(g1(x)) -> h1(a)
g1(h1(x)) -> g1(x)
h1(h1(x)) -> x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G1(h1(x)) -> G1(x)
The TRS R consists of the following rules:
f2(a, h1(x)) -> f2(g1(x), h1(x))
h1(g1(x)) -> h1(a)
g1(h1(x)) -> g1(x)
h1(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(h1(x)) -> G1(x)
Used argument filtering: G1(x1) = x1
h1(x1) = h1(x1)
Used ordering: 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, h1(x)) -> f2(g1(x), h1(x))
h1(g1(x)) -> h1(a)
g1(h1(x)) -> g1(x)
h1(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.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F2(a, h1(x)) -> F2(g1(x), h1(x))
The TRS R consists of the following rules:
f2(a, h1(x)) -> f2(g1(x), h1(x))
h1(g1(x)) -> h1(a)
g1(h1(x)) -> g1(x)
h1(h1(x)) -> x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.