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:
f2(x, a1(b1(y))) -> f2(c1(d1(x)), y)
f2(c1(x), y) -> f2(x, a1(y))
f2(d1(x), y) -> f2(x, b1(y))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(x, a1(b1(y))) -> f2(c1(d1(x)), y)
f2(c1(x), y) -> f2(x, a1(y))
f2(d1(x), y) -> f2(x, b1(y))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F2(d1(x), y) -> F2(x, b1(y))
F2(x, a1(b1(y))) -> F2(c1(d1(x)), y)
F2(c1(x), y) -> F2(x, a1(y))
The TRS R consists of the following rules:
f2(x, a1(b1(y))) -> f2(c1(d1(x)), y)
f2(c1(x), y) -> f2(x, a1(y))
f2(d1(x), y) -> f2(x, b1(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:
F2(d1(x), y) -> F2(x, b1(y))
F2(x, a1(b1(y))) -> F2(c1(d1(x)), y)
F2(c1(x), y) -> F2(x, a1(y))
The TRS R consists of the following rules:
f2(x, a1(b1(y))) -> f2(c1(d1(x)), y)
f2(c1(x), y) -> f2(x, a1(y))
f2(d1(x), y) -> f2(x, b1(y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.