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.