Termination w.r.t. Q proof of TRS/Zantemaz18.trs

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:

f(x, a(b(y))) → a(f(a(b(x)), y))
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

f(x, a(b(y))) → a(f(a(b(x)), y))
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(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:

F(b(x), y) → F(x, b(y))
F(a(x), y) → F(x, a(y))
F(x, a(b(y))) → F(a(b(x)), y)

The TRS R consists of the following rules:

f(x, a(b(y))) → a(f(a(b(x)), y))
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

Q DP problem:
The TRS P consists of the following rules:

F(b(x), y) → F(x, b(y))
F(a(x), y) → F(x, a(y))
F(x, a(b(y))) → F(a(b(x)), y)

The TRS R consists of the following rules:

f(x, a(b(y))) → a(f(a(b(x)), y))
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

F(b(x), y) → F(x, b(y))
F(a(x), y) → F(x, a(y))
F(x, a(b(y))) → F(a(b(x)), y)

The TRS R consists of the following rules:

f(x, a(b(y))) → a(f(a(b(x)), y))
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.