Termination w.r.t. Q proof of TRS/Zantemaz24.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))) → f(a(b(x)), y)
f(x, b(c(y))) → f(b(c(x)), y)
f(x, c(a(y))) → f(c(a(x)), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
f(c(x), y) → f(x, c(y))

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

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

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

Q is empty.

We have applied [15,7] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

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

The signature Sigma is {f}

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

f(x0, a(b(x1)))
f(x0, b(c(x1)))
f(x0, c(a(x1)))
f(a(x0), x1)
f(b(x0), x1)
f(c(x0), x1)

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, a(b(x1)))
f(x0, b(c(x1)))
f(x0, c(a(x1)))
f(a(x0), x1)
f(b(x0), x1)
f(c(x0), x1)

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, a(b(x1)))
f(x0, b(c(x1)))
f(x0, c(a(x1)))
f(a(x0), x1)
f(b(x0), x1)
f(c(x0), x1)

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, a(b(x1)))
f(x0, b(c(x1)))
f(x0, c(a(x1)))
f(a(x0), x1)
f(b(x0), x1)
f(c(x0), x1)

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