Termination w.r.t. Q proof of TRS/Cimefilliatre2.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
foldB(t, 0) → t
foldB(t, s(n)) → f(foldB(t, n), B)
foldC(t, 0) → t
foldC(t, s(n)) → f(foldC(t, n), C)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, s(c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) → foldC(triple(a, b, 0), 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
foldB(t, 0) → t
foldB(t, s(n)) → f(foldB(t, n), B)
foldC(t, 0) → t
foldC(t, s(n)) → f(foldC(t, n), C)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, s(c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) → foldC(triple(a, b, 0), 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:

F'(triple(a, b, c), A) → FOLDB(triple(s(a), 0, c), b)
F(t, x) → F'(t, g(x))
FOLDC(t, s(n)) → F(foldC(t, n), C)
F(t, x) → G(x)
FOLDB(t, s(n)) → FOLDB(t, n)
F''(triple(a, b, c)) → FOLDC(triple(a, b, 0), c)
FOLDC(t, s(n)) → FOLDC(t, n)
F'(triple(a, b, c), A) → F''(foldB(triple(s(a), 0, c), b))
F'(triple(a, b, c), B) → F(triple(a, b, c), A)
FOLDB(t, s(n)) → F(foldB(t, n), B)

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
foldB(t, 0) → t
foldB(t, s(n)) → f(foldB(t, n), B)
foldC(t, 0) → t
foldC(t, s(n)) → f(foldC(t, n), C)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, s(c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) → foldC(triple(a, b, 0), 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:

F'(triple(a, b, c), A) → FOLDB(triple(s(a), 0, c), b)
F(t, x) → F'(t, g(x))
FOLDC(t, s(n)) → F(foldC(t, n), C)
F(t, x) → G(x)
FOLDB(t, s(n)) → FOLDB(t, n)
F''(triple(a, b, c)) → FOLDC(triple(a, b, 0), c)
FOLDC(t, s(n)) → FOLDC(t, n)
F'(triple(a, b, c), A) → F''(foldB(triple(s(a), 0, c), b))
F'(triple(a, b, c), B) → F(triple(a, b, c), A)
FOLDB(t, s(n)) → F(foldB(t, n), B)

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
foldB(t, 0) → t
foldB(t, s(n)) → f(foldB(t, n), B)
foldC(t, 0) → t
foldC(t, s(n)) → f(foldC(t, n), C)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, s(c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) → foldC(triple(a, b, 0), 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:

F'(triple(a, b, c), A) → FOLDB(triple(s(a), 0, c), b)
F(t, x) → F'(t, g(x))
FOLDC(t, s(n)) → F(foldC(t, n), C)
FOLDB(t, s(n)) → FOLDB(t, n)
F(t, x) → G(x)
F''(triple(a, b, c)) → FOLDC(triple(a, b, 0), c)
FOLDC(t, s(n)) → FOLDC(t, n)
F'(triple(a, b, c), B) → F(triple(a, b, c), A)
F'(triple(a, b, c), A) → F''(foldB(triple(s(a), 0, c), b))
FOLDB(t, s(n)) → F(foldB(t, n), B)

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
foldB(t, 0) → t
foldB(t, s(n)) → f(foldB(t, n), B)
foldC(t, 0) → t
foldC(t, s(n)) → f(foldC(t, n), C)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, s(c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) → foldC(triple(a, b, 0), 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:

F'(triple(a, b, c), A) → FOLDB(triple(s(a), 0, c), b)
F(t, x) → F'(t, g(x))
FOLDC(t, s(n)) → F(foldC(t, n), C)
FOLDB(t, s(n)) → FOLDB(t, n)
F''(triple(a, b, c)) → FOLDC(triple(a, b, 0), c)
FOLDC(t, s(n)) → FOLDC(t, n)
F'(triple(a, b, c), A) → F''(foldB(triple(s(a), 0, c), b))
F'(triple(a, b, c), B) → F(triple(a, b, c), A)
FOLDB(t, s(n)) → F(foldB(t, n), B)

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
foldB(t, 0) → t
foldB(t, s(n)) → f(foldB(t, n), B)
foldC(t, 0) → t
foldC(t, s(n)) → f(foldC(t, n), C)
f(t, x) → f'(t, g(x))
f'(triple(a, b, c), C) → triple(a, b, s(c))
f'(triple(a, b, c), B) → f(triple(a, b, c), A)
f'(triple(a, b, c), A) → f''(foldB(triple(s(a), 0, c), b))
f''(triple(a, b, c)) → foldC(triple(a, b, 0), c)

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