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(c1(f1(x)))
f1(f1(x)) -> f1(d1(f1(x)))
g1(c1(x)) -> x
g1(d1(x)) -> x
g1(c1(h1(0))) -> g1(d1(1))
g1(c1(1)) -> g1(d1(h1(0)))
g1(h1(x)) -> g1(x)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(f1(x)) -> f1(c1(f1(x)))
f1(f1(x)) -> f1(d1(f1(x)))
g1(c1(x)) -> x
g1(d1(x)) -> x
g1(c1(h1(0))) -> g1(d1(1))
g1(c1(1)) -> g1(d1(h1(0)))
g1(h1(x)) -> g1(x)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F1(f1(x)) -> F1(c1(f1(x)))
F1(f1(x)) -> F1(d1(f1(x)))
G1(c1(h1(0))) -> G1(d1(1))
G1(h1(x)) -> G1(x)
G1(c1(1)) -> G1(d1(h1(0)))
The TRS R consists of the following rules:
f1(f1(x)) -> f1(c1(f1(x)))
f1(f1(x)) -> f1(d1(f1(x)))
g1(c1(x)) -> x
g1(d1(x)) -> x
g1(c1(h1(0))) -> g1(d1(1))
g1(c1(1)) -> g1(d1(h1(0)))
g1(h1(x)) -> g1(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:
F1(f1(x)) -> F1(c1(f1(x)))
F1(f1(x)) -> F1(d1(f1(x)))
G1(c1(h1(0))) -> G1(d1(1))
G1(h1(x)) -> G1(x)
G1(c1(1)) -> G1(d1(h1(0)))
The TRS R consists of the following rules:
f1(f1(x)) -> f1(c1(f1(x)))
f1(f1(x)) -> f1(d1(f1(x)))
g1(c1(x)) -> x
g1(d1(x)) -> x
g1(c1(h1(0))) -> g1(d1(1))
g1(c1(1)) -> g1(d1(h1(0)))
g1(h1(x)) -> g1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
G1(h1(x)) -> G1(x)
The TRS R consists of the following rules:
f1(f1(x)) -> f1(c1(f1(x)))
f1(f1(x)) -> f1(d1(f1(x)))
g1(c1(x)) -> x
g1(d1(x)) -> x
g1(c1(h1(0))) -> g1(d1(1))
g1(c1(1)) -> g1(d1(h1(0)))
g1(h1(x)) -> g1(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
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f1(f1(x)) -> f1(c1(f1(x)))
f1(f1(x)) -> f1(d1(f1(x)))
g1(c1(x)) -> x
g1(d1(x)) -> x
g1(c1(h1(0))) -> g1(d1(1))
g1(c1(1)) -> g1(d1(h1(0)))
g1(h1(x)) -> g1(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.