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(a1(a1(x)), y)
f2(x, b1(a1(y))) -> f2(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(b1(y))) -> f2(a1(a1(x)), y)
f2(x, b1(a1(y))) -> f2(b1(b1(x)), y)
f2(a1(x), y) -> f2(x, a1(y))
f2(b1(x), y) -> f2(x, b1(y))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F2(x, a1(b1(y))) -> F2(a1(a1(x)), y)
F2(b1(x), y) -> F2(x, b1(y))
F2(a1(x), y) -> F2(x, a1(y))
F2(x, b1(a1(y))) -> F2(b1(b1(x)), y)

The TRS R consists of the following rules:

f2(x, a1(b1(y))) -> f2(a1(a1(x)), y)
f2(x, b1(a1(y))) -> f2(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(x, a1(b1(y))) -> F2(a1(a1(x)), y)
F2(b1(x), y) -> F2(x, b1(y))
F2(a1(x), y) -> F2(x, a1(y))
F2(x, b1(a1(y))) -> F2(b1(b1(x)), y)

The TRS R consists of the following rules:

f2(x, a1(b1(y))) -> f2(a1(a1(x)), y)
f2(x, b1(a1(y))) -> f2(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.