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(a) -> f1(c1(a))
f1(c1(X)) -> X
f1(c1(a)) -> f1(d1(b))
f1(a) -> f1(d1(a))
f1(d1(X)) -> X
f1(c1(b)) -> f1(d1(a))
e1(g1(X)) -> e1(X)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(a) -> f1(c1(a))
f1(c1(X)) -> X
f1(c1(a)) -> f1(d1(b))
f1(a) -> f1(d1(a))
f1(d1(X)) -> X
f1(c1(b)) -> f1(d1(a))
e1(g1(X)) -> e1(X)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F1(c1(a)) -> F1(d1(b))
F1(c1(b)) -> F1(d1(a))
F1(a) -> F1(d1(a))
E1(g1(X)) -> E1(X)
F1(a) -> F1(c1(a))
The TRS R consists of the following rules:
f1(a) -> f1(c1(a))
f1(c1(X)) -> X
f1(c1(a)) -> f1(d1(b))
f1(a) -> f1(d1(a))
f1(d1(X)) -> X
f1(c1(b)) -> f1(d1(a))
e1(g1(X)) -> e1(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(c1(a)) -> F1(d1(b))
F1(c1(b)) -> F1(d1(a))
F1(a) -> F1(d1(a))
E1(g1(X)) -> E1(X)
F1(a) -> F1(c1(a))
The TRS R consists of the following rules:
f1(a) -> f1(c1(a))
f1(c1(X)) -> X
f1(c1(a)) -> f1(d1(b))
f1(a) -> f1(d1(a))
f1(d1(X)) -> X
f1(c1(b)) -> f1(d1(a))
e1(g1(X)) -> e1(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:
E1(g1(X)) -> E1(X)
The TRS R consists of the following rules:
f1(a) -> f1(c1(a))
f1(c1(X)) -> X
f1(c1(a)) -> f1(d1(b))
f1(a) -> f1(d1(a))
f1(d1(X)) -> X
f1(c1(b)) -> f1(d1(a))
e1(g1(X)) -> e1(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.
E1(g1(X)) -> E1(X)
Used argument filtering: E1(x1) = x1
g1(x1) = g1(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(a) -> f1(c1(a))
f1(c1(X)) -> X
f1(c1(a)) -> f1(d1(b))
f1(a) -> f1(d1(a))
f1(d1(X)) -> X
f1(c1(b)) -> f1(d1(a))
e1(g1(X)) -> e1(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.