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(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(x) -> x
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(x) -> x
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F1(f1(x)) -> F1(d1(f1(x)))
F1(c2(s1(x), y)) -> F1(c2(x, s1(y)))
G1(c2(x, s1(y))) -> G1(c2(s1(x), y))
The TRS R consists of the following rules:
g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(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:
F1(f1(x)) -> F1(d1(f1(x)))
F1(c2(s1(x), y)) -> F1(c2(x, s1(y)))
G1(c2(x, s1(y))) -> G1(c2(s1(x), y))
The TRS R consists of the following rules:
g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F1(c2(s1(x), y)) -> F1(c2(x, s1(y)))
The TRS R consists of the following rules:
g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(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.
F1(c2(s1(x), y)) -> F1(c2(x, s1(y)))
Used argument filtering: F1(x1) = x1
c2(x1, x2) = x1
s1(x1) = s1(x1)
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:
g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(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
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
G1(c2(x, s1(y))) -> G1(c2(s1(x), y))
The TRS R consists of the following rules:
g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(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(c2(x, s1(y))) -> G1(c2(s1(x), y))
Used argument filtering: G1(x1) = x1
c2(x1, x2) = x2
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(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.