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