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:
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(c2(s1(x), s1(y))) -> g1(c2(x, y))
g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
g1(c2(s1(x), s1(y))) -> f1(c2(x, y))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(c2(s1(x), s1(y))) -> g1(c2(x, y))
g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
g1(c2(s1(x), s1(y))) -> f1(c2(x, y))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F1(c2(s1(x), s1(y))) -> G1(c2(x, y))
F1(c2(s1(x), y)) -> F1(c2(x, s1(y)))
G1(c2(x, s1(y))) -> G1(c2(s1(x), y))
G1(c2(s1(x), s1(y))) -> F1(c2(x, y))
The TRS R consists of the following rules:
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(c2(s1(x), s1(y))) -> g1(c2(x, y))
g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
g1(c2(s1(x), s1(y))) -> f1(c2(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F1(c2(s1(x), s1(y))) -> G1(c2(x, y))
F1(c2(s1(x), y)) -> F1(c2(x, s1(y)))
G1(c2(x, s1(y))) -> G1(c2(s1(x), y))
G1(c2(s1(x), s1(y))) -> F1(c2(x, y))
The TRS R consists of the following rules:
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(c2(s1(x), s1(y))) -> g1(c2(x, y))
g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
g1(c2(s1(x), s1(y))) -> f1(c2(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.