Termination w.r.t. Q proof of TRS/Cimefilliatre.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:

g(A) → A
g(B) → A
g(B) → B
g(C) → A
g(C) → B
g(C) → C
foldf(x, nil) → x
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, cons(C, c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) → foldf(triple(a, b, nil), c)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

g(A) → A
g(B) → A
g(B) → B
g(C) → A
g(C) → B
g(C) → C
foldf(x, nil) → x
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, cons(C, c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) → foldf(triple(a, b, nil), c)

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:

FOLDF(x, cons(y, z)) → FOLDF(x, z)
FOLDF(x, cons(y, z)) → F(foldf(x, z), y)
F'(triple(a, b, c), A) → F''(foldf(triple(cons(A, a), nil, c), b))
F(t, x) → F'(t, g(x))
F(t, x) → G(x)
F'(triple(a, b, c), A) → FOLDF(triple(cons(A, a), nil, c), b)
F''(triple(a, b, c)) → FOLDF(triple(a, b, nil), c)
F'(triple(a, b, c), B) → F(triple(a, b, c), A)

The TRS R consists of the following rules:

g(A) → A
g(B) → A
g(B) → B
g(C) → A
g(C) → B
g(C) → C
foldf(x, nil) → x
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, cons(C, c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) → foldf(triple(a, b, nil), c)

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:

FOLDF(x, cons(y, z)) → FOLDF(x, z)
FOLDF(x, cons(y, z)) → F(foldf(x, z), y)
F'(triple(a, b, c), A) → F''(foldf(triple(cons(A, a), nil, c), b))
F(t, x) → F'(t, g(x))
F(t, x) → G(x)
F'(triple(a, b, c), A) → FOLDF(triple(cons(A, a), nil, c), b)
F''(triple(a, b, c)) → FOLDF(triple(a, b, nil), c)
F'(triple(a, b, c), B) → F(triple(a, b, c), A)

The TRS R consists of the following rules:

g(A) → A
g(B) → A
g(B) → B
g(C) → A
g(C) → B
g(C) → C
foldf(x, nil) → x
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, cons(C, c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) → foldf(triple(a, b, nil), c)

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

          ↳ DependencyGraphProof

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

FOLDF(x, cons(y, z)) → FOLDF(x, z)
FOLDF(x, cons(y, z)) → F(foldf(x, z), y)
F(t, x) → F'(t, g(x))
F'(triple(a, b, c), A) → F''(foldf(triple(cons(A, a), nil, c), b))
F(t, x) → G(x)
F'(triple(a, b, c), A) → FOLDF(triple(cons(A, a), nil, c), b)
F'(triple(a, b, c), B) → F(triple(a, b, c), A)
F''(triple(a, b, c)) → FOLDF(triple(a, b, nil), c)

The TRS R consists of the following rules:

g(A) → A
g(B) → A
g(B) → B
g(C) → A
g(C) → B
g(C) → C
foldf(x, nil) → x
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, cons(C, c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) → foldf(triple(a, b, nil), c)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

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

FOLDF(x, cons(y, z)) → FOLDF(x, z)
FOLDF(x, cons(y, z)) → F(foldf(x, z), y)
F'(triple(a, b, c), A) → F''(foldf(triple(cons(A, a), nil, c), b))
F(t, x) → F'(t, g(x))
F'(triple(a, b, c), A) → FOLDF(triple(cons(A, a), nil, c), b)
F''(triple(a, b, c)) → FOLDF(triple(a, b, nil), c)
F'(triple(a, b, c), B) → F(triple(a, b, c), A)

The TRS R consists of the following rules:

g(A) → A
g(B) → A
g(B) → B
g(C) → A
g(C) → B
g(C) → C
foldf(x, nil) → x
foldf(x, cons(y, z)) → f(foldf(x, z), y)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, cons(C, c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldf(triple(cons(A, a), nil, c), b))
f''(triple(a, b, c)) → foldf(triple(a, b, nil), c)

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