Termination w.r.t. Q proof of TRS/AG01#3.49.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(c(s(x), y)) → f(c(x, s(y)))
f(c(s(x), s(y))) → g(c(x, y))
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

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

f(c(s(x), y)) → f(c(x, s(y)))
f(c(s(x), s(y))) → g(c(x, y))
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, 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(c(s(x), y)) → f(c(x, s(y)))
f(c(s(x), s(y))) → g(c(x, y))
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))

The signature Sigma is {f, g}

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

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

f(c(s(x), y)) → f(c(x, s(y)))
f(c(s(x), s(y))) → g(c(x, y))
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))

The set Q consists of the following terms:

f(c(s(x0), x1))
g(c(x0, s(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(c(s(x), s(y))) → G(c(x, y))
F(c(s(x), y)) → F(c(x, s(y)))
G(c(x, s(y))) → G(c(s(x), y))
G(c(s(x), s(y))) → F(c(x, y))

The TRS R consists of the following rules:

f(c(s(x), y)) → f(c(x, s(y)))
f(c(s(x), s(y))) → g(c(x, y))
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))

The set Q consists of the following terms:

f(c(s(x0), x1))
g(c(x0, s(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(c(s(x), s(y))) → G(c(x, y))
F(c(s(x), y)) → F(c(x, s(y)))
G(c(x, s(y))) → G(c(s(x), y))
G(c(s(x), s(y))) → F(c(x, y))

The TRS R consists of the following rules:

f(c(s(x), y)) → f(c(x, s(y)))
f(c(s(x), s(y))) → g(c(x, y))
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))

The set Q consists of the following terms:

f(c(s(x0), x1))
g(c(x0, s(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(c(s(x), s(y))) → G(c(x, y))
F(c(s(x), y)) → F(c(x, s(y)))
G(c(x, s(y))) → G(c(s(x), y))
G(c(s(x), s(y))) → F(c(x, y))

The TRS R consists of the following rules:

f(c(s(x), y)) → f(c(x, s(y)))
f(c(s(x), s(y))) → g(c(x, y))
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))

The set Q consists of the following terms:

f(c(s(x0), x1))
g(c(x0, s(x1)))

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