Termination w.r.t. Q proof of TRS/Cimefilliatre3.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:

g₁(A) -> A
g₁(B) -> A
g₁(B) -> B
g₁(C) -> A
g₁(C) -> B
g₁(C) -> C
foldB₂(t, 0) -> t
foldB₂(t, s₁(n)) -> f₂(foldB₂(t, n), B)
foldC₂(t, 0) -> t
foldC₂(t, s₁(n)) -> f₂(foldC₂(t, n), C)
f₂(t, x) -> f'₂(t, g₁(x))
f'₂(triple₃(a, b, c), C) -> triple₃(a, b, s₁(c))
f'₂(triple₃(a, b, c), B) -> f₂(triple₃(a, b, c), A)
f'₂(triple₃(a, b, c), A) -> f''₁(foldB₂(triple₃(s₁(a), 0, c), b))
f''₁(triple₃(a, b, c)) -> foldC₂(triple₃(a, b, 0), c)
fold₃(t, x, 0) -> t
fold₃(t, x, s₁(n)) -> f₂(fold₃(t, x, n), x)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

g₁(A) -> A
g₁(B) -> A
g₁(B) -> B
g₁(C) -> A
g₁(C) -> B
g₁(C) -> C
foldB₂(t, 0) -> t
foldB₂(t, s₁(n)) -> f₂(foldB₂(t, n), B)
foldC₂(t, 0) -> t
foldC₂(t, s₁(n)) -> f₂(foldC₂(t, n), C)
f₂(t, x) -> f'₂(t, g₁(x))
f'₂(triple₃(a, b, c), C) -> triple₃(a, b, s₁(c))
f'₂(triple₃(a, b, c), B) -> f₂(triple₃(a, b, c), A)
f'₂(triple₃(a, b, c), A) -> f''₁(foldB₂(triple₃(s₁(a), 0, c), b))
f''₁(triple₃(a, b, c)) -> foldC₂(triple₃(a, b, 0), c)
fold₃(t, x, 0) -> t
fold₃(t, x, s₁(n)) -> f₂(fold₃(t, x, n), x)

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'₂(triple₃(a, b, c), A) -> FOLDB₂(triple₃(s₁(a), 0, c), b)
FOLD₃(t, x, s₁(n)) -> FOLD₃(t, x, n)
FOLD₃(t, x, s₁(n)) -> F₂(fold₃(t, x, n), x)
F₂(t, x) -> F'₂(t, g₁(x))
FOLDC₂(t, s₁(n)) -> F₂(foldC₂(t, n), C)
F₂(t, x) -> G₁(x)
FOLDB₂(t, s₁(n)) -> FOLDB₂(t, n)
F''₁(triple₃(a, b, c)) -> FOLDC₂(triple₃(a, b, 0), c)
FOLDC₂(t, s₁(n)) -> FOLDC₂(t, n)
F'₂(triple₃(a, b, c), A) -> F''₁(foldB₂(triple₃(s₁(a), 0, c), b))
F'₂(triple₃(a, b, c), B) -> F₂(triple₃(a, b, c), A)
FOLDB₂(t, s₁(n)) -> F₂(foldB₂(t, n), B)

The TRS R consists of the following rules:

g₁(A) -> A
g₁(B) -> A
g₁(B) -> B
g₁(C) -> A
g₁(C) -> B
g₁(C) -> C
foldB₂(t, 0) -> t
foldB₂(t, s₁(n)) -> f₂(foldB₂(t, n), B)
foldC₂(t, 0) -> t
foldC₂(t, s₁(n)) -> f₂(foldC₂(t, n), C)
f₂(t, x) -> f'₂(t, g₁(x))
f'₂(triple₃(a, b, c), C) -> triple₃(a, b, s₁(c))
f'₂(triple₃(a, b, c), B) -> f₂(triple₃(a, b, c), A)
f'₂(triple₃(a, b, c), A) -> f''₁(foldB₂(triple₃(s₁(a), 0, c), b))
f''₁(triple₃(a, b, c)) -> foldC₂(triple₃(a, b, 0), c)
fold₃(t, x, 0) -> t
fold₃(t, x, s₁(n)) -> f₂(fold₃(t, x, n), x)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

F'₂(triple₃(a, b, c), A) -> FOLDB₂(triple₃(s₁(a), 0, c), b)
FOLD₃(t, x, s₁(n)) -> FOLD₃(t, x, n)
FOLD₃(t, x, s₁(n)) -> F₂(fold₃(t, x, n), x)
F₂(t, x) -> F'₂(t, g₁(x))
FOLDC₂(t, s₁(n)) -> F₂(foldC₂(t, n), C)
F₂(t, x) -> G₁(x)
FOLDB₂(t, s₁(n)) -> FOLDB₂(t, n)
F''₁(triple₃(a, b, c)) -> FOLDC₂(triple₃(a, b, 0), c)
FOLDC₂(t, s₁(n)) -> FOLDC₂(t, n)
F'₂(triple₃(a, b, c), A) -> F''₁(foldB₂(triple₃(s₁(a), 0, c), b))
F'₂(triple₃(a, b, c), B) -> F₂(triple₃(a, b, c), A)
FOLDB₂(t, s₁(n)) -> F₂(foldB₂(t, n), B)

The TRS R consists of the following rules:

g₁(A) -> A
g₁(B) -> A
g₁(B) -> B
g₁(C) -> A
g₁(C) -> B
g₁(C) -> C
foldB₂(t, 0) -> t
foldB₂(t, s₁(n)) -> f₂(foldB₂(t, n), B)
foldC₂(t, 0) -> t
foldC₂(t, s₁(n)) -> f₂(foldC₂(t, n), C)
f₂(t, x) -> f'₂(t, g₁(x))
f'₂(triple₃(a, b, c), C) -> triple₃(a, b, s₁(c))
f'₂(triple₃(a, b, c), B) -> f₂(triple₃(a, b, c), A)
f'₂(triple₃(a, b, c), A) -> f''₁(foldB₂(triple₃(s₁(a), 0, c), b))
f''₁(triple₃(a, b, c)) -> foldC₂(triple₃(a, b, 0), c)
fold₃(t, x, 0) -> t
fold₃(t, x, s₁(n)) -> f₂(fold₃(t, x, n), x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

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

F'₂(triple₃(a, b, c), A) -> FOLDB₂(triple₃(s₁(a), 0, c), b)
F₂(t, x) -> F'₂(t, g₁(x))
FOLDC₂(t, s₁(n)) -> F₂(foldC₂(t, n), C)
FOLDB₂(t, s₁(n)) -> FOLDB₂(t, n)
F''₁(triple₃(a, b, c)) -> FOLDC₂(triple₃(a, b, 0), c)
FOLDC₂(t, s₁(n)) -> FOLDC₂(t, n)
F'₂(triple₃(a, b, c), A) -> F''₁(foldB₂(triple₃(s₁(a), 0, c), b))
F'₂(triple₃(a, b, c), B) -> F₂(triple₃(a, b, c), A)
FOLDB₂(t, s₁(n)) -> F₂(foldB₂(t, n), B)

The TRS R consists of the following rules:

g₁(A) -> A
g₁(B) -> A
g₁(B) -> B
g₁(C) -> A
g₁(C) -> B
g₁(C) -> C
foldB₂(t, 0) -> t
foldB₂(t, s₁(n)) -> f₂(foldB₂(t, n), B)
foldC₂(t, 0) -> t
foldC₂(t, s₁(n)) -> f₂(foldC₂(t, n), C)
f₂(t, x) -> f'₂(t, g₁(x))
f'₂(triple₃(a, b, c), C) -> triple₃(a, b, s₁(c))
f'₂(triple₃(a, b, c), B) -> f₂(triple₃(a, b, c), A)
f'₂(triple₃(a, b, c), A) -> f''₁(foldB₂(triple₃(s₁(a), 0, c), b))
f''₁(triple₃(a, b, c)) -> foldC₂(triple₃(a, b, 0), c)
fold₃(t, x, 0) -> t
fold₃(t, x, s₁(n)) -> f₂(fold₃(t, x, n), x)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

FOLD₃(t, x, s₁(n)) -> FOLD₃(t, x, n)

The TRS R consists of the following rules:

g₁(A) -> A
g₁(B) -> A
g₁(B) -> B
g₁(C) -> A
g₁(C) -> B
g₁(C) -> C
foldB₂(t, 0) -> t
foldB₂(t, s₁(n)) -> f₂(foldB₂(t, n), B)
foldC₂(t, 0) -> t
foldC₂(t, s₁(n)) -> f₂(foldC₂(t, n), C)
f₂(t, x) -> f'₂(t, g₁(x))
f'₂(triple₃(a, b, c), C) -> triple₃(a, b, s₁(c))
f'₂(triple₃(a, b, c), B) -> f₂(triple₃(a, b, c), A)
f'₂(triple₃(a, b, c), A) -> f''₁(foldB₂(triple₃(s₁(a), 0, c), b))
f''₁(triple₃(a, b, c)) -> foldC₂(triple₃(a, b, 0), c)
fold₃(t, x, 0) -> t
fold₃(t, x, s₁(n)) -> f₂(fold₃(t, x, n), x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

FOLD₃(t, x, s₁(n)) -> FOLD₃(t, x, n)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(FOLD₃(x₁, x₂, x₃)) = x₃
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

g₁(A) -> A
g₁(B) -> A
g₁(B) -> B
g₁(C) -> A
g₁(C) -> B
g₁(C) -> C
foldB₂(t, 0) -> t
foldB₂(t, s₁(n)) -> f₂(foldB₂(t, n), B)
foldC₂(t, 0) -> t
foldC₂(t, s₁(n)) -> f₂(foldC₂(t, n), C)
f₂(t, x) -> f'₂(t, g₁(x))
f'₂(triple₃(a, b, c), C) -> triple₃(a, b, s₁(c))
f'₂(triple₃(a, b, c), B) -> f₂(triple₃(a, b, c), A)
f'₂(triple₃(a, b, c), A) -> f''₁(foldB₂(triple₃(s₁(a), 0, c), b))
f''₁(triple₃(a, b, c)) -> foldC₂(triple₃(a, b, 0), c)
fold₃(t, x, 0) -> t
fold₃(t, x, s₁(n)) -> f₂(fold₃(t, x, n), x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.