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

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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
foldf₂(x, nil) -> x
foldf₂(x, cons₂(y, z)) -> f₂(foldf₂(x, z), y)
f₂(t, x) -> f'₂(t, g₁(x))
f'₂(triple₃(a, b, c), C) -> triple₃(a, b, cons₂(C, c))
f'₂(triple₃(a, b, c), B) -> f₂(triple₃(a, b, c), A)
f'₂(triple₃(a, b, c), A) -> f''₁(foldf₂(triple₃(cons₂(A, a), nil, c), b))
f''₁(triple₃(a, b, c)) -> foldf₂(triple₃(a, b, nil), c)

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
foldf₂(x, nil) -> x
foldf₂(x, cons₂(y, z)) -> f₂(foldf₂(x, z), y)
f₂(t, x) -> f'₂(t, g₁(x))
f'₂(triple₃(a, b, c), C) -> triple₃(a, b, cons₂(C, c))
f'₂(triple₃(a, b, c), B) -> f₂(triple₃(a, b, c), A)
f'₂(triple₃(a, b, c), A) -> f''₁(foldf₂(triple₃(cons₂(A, a), nil, c), b))
f''₁(triple₃(a, b, c)) -> foldf₂(triple₃(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:

FOLDF₂(x, cons₂(y, z)) -> FOLDF₂(x, z)
FOLDF₂(x, cons₂(y, z)) -> F₂(foldf₂(x, z), y)
F'₂(triple₃(a, b, c), A) -> F''₁(foldf₂(triple₃(cons₂(A, a), nil, c), b))
F₂(t, x) -> F'₂(t, g₁(x))
F₂(t, x) -> G₁(x)
F'₂(triple₃(a, b, c), A) -> FOLDF₂(triple₃(cons₂(A, a), nil, c), b)
F''₁(triple₃(a, b, c)) -> FOLDF₂(triple₃(a, b, nil), c)
F'₂(triple₃(a, b, c), B) -> F₂(triple₃(a, b, c), A)

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
foldf₂(x, nil) -> x
foldf₂(x, cons₂(y, z)) -> f₂(foldf₂(x, z), y)
f₂(t, x) -> f'₂(t, g₁(x))
f'₂(triple₃(a, b, c), C) -> triple₃(a, b, cons₂(C, c))
f'₂(triple₃(a, b, c), B) -> f₂(triple₃(a, b, c), A)
f'₂(triple₃(a, b, c), A) -> f''₁(foldf₂(triple₃(cons₂(A, a), nil, c), b))
f''₁(triple₃(a, b, c)) -> foldf₂(triple₃(a, b, nil), c)

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:

FOLDF₂(x, cons₂(y, z)) -> FOLDF₂(x, z)
FOLDF₂(x, cons₂(y, z)) -> F₂(foldf₂(x, z), y)
F'₂(triple₃(a, b, c), A) -> F''₁(foldf₂(triple₃(cons₂(A, a), nil, c), b))
F₂(t, x) -> F'₂(t, g₁(x))
F₂(t, x) -> G₁(x)
F'₂(triple₃(a, b, c), A) -> FOLDF₂(triple₃(cons₂(A, a), nil, c), b)
F''₁(triple₃(a, b, c)) -> FOLDF₂(triple₃(a, b, nil), c)
F'₂(triple₃(a, b, c), B) -> F₂(triple₃(a, b, c), A)

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
foldf₂(x, nil) -> x
foldf₂(x, cons₂(y, z)) -> f₂(foldf₂(x, z), y)
f₂(t, x) -> f'₂(t, g₁(x))
f'₂(triple₃(a, b, c), C) -> triple₃(a, b, cons₂(C, c))
f'₂(triple₃(a, b, c), B) -> f₂(triple₃(a, b, c), A)
f'₂(triple₃(a, b, c), A) -> f''₁(foldf₂(triple₃(cons₂(A, a), nil, c), b))
f''₁(triple₃(a, b, c)) -> foldf₂(triple₃(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

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

FOLDF₂(x, cons₂(y, z)) -> FOLDF₂(x, z)
FOLDF₂(x, cons₂(y, z)) -> F₂(foldf₂(x, z), y)
F'₂(triple₃(a, b, c), A) -> F''₁(foldf₂(triple₃(cons₂(A, a), nil, c), b))
F₂(t, x) -> F'₂(t, g₁(x))
F'₂(triple₃(a, b, c), A) -> FOLDF₂(triple₃(cons₂(A, a), nil, c), b)
F''₁(triple₃(a, b, c)) -> FOLDF₂(triple₃(a, b, nil), c)
F'₂(triple₃(a, b, c), B) -> F₂(triple₃(a, b, c), A)

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
foldf₂(x, nil) -> x
foldf₂(x, cons₂(y, z)) -> f₂(foldf₂(x, z), y)
f₂(t, x) -> f'₂(t, g₁(x))
f'₂(triple₃(a, b, c), C) -> triple₃(a, b, cons₂(C, c))
f'₂(triple₃(a, b, c), B) -> f₂(triple₃(a, b, c), A)
f'₂(triple₃(a, b, c), A) -> f''₁(foldf₂(triple₃(cons₂(A, a), nil, c), b))
f''₁(triple₃(a, b, c)) -> foldf₂(triple₃(a, b, nil), c)

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.

FOLDF₂(x, cons₂(y, z)) -> FOLDF₂(x, z)
FOLDF₂(x, cons₂(y, z)) -> F₂(foldf₂(x, z), y)

The remaining pairs can at least be oriented weakly.

F'₂(triple₃(a, b, c), A) -> F''₁(foldf₂(triple₃(cons₂(A, a), nil, c), b))
F₂(t, x) -> F'₂(t, g₁(x))
F'₂(triple₃(a, b, c), A) -> FOLDF₂(triple₃(cons₂(A, a), nil, c), b)
F''₁(triple₃(a, b, c)) -> FOLDF₂(triple₃(a, b, nil), c)
F'₂(triple₃(a, b, c), B) -> F₂(triple₃(a, b, c), A)

Used ordering: Polynomial interpretation [21]:

POL(A) = 0
POL(B) = 0
POL(C) = 0
POL(F₂(x₁, x₂)) = 1 + x₁ + x₂
POL(F'₂(x₁, x₂)) = 1 + x₁ + x₂
POL(F''₁(x₁)) = 1 + x₁
POL(FOLDF₂(x₁, x₂)) = 1 + x₁ + x₂
POL(cons₂(x₁, x₂)) = 1 + x₁ + x₂
POL(f₂(x₁, x₂)) = 1 + x₁
POL(f'₂(x₁, x₂)) = 1 + x₁
POL(f''₁(x₁)) = x₁
POL(foldf₂(x₁, x₂)) = x₁ + x₂
POL(g₁(x₁)) = x₁
POL(nil) = 0
POL(triple₃(x₁, x₂, x₃)) = x₂ + x₃

The following usable rules [14] were oriented:

g₁(C) -> B
g₁(C) -> A
f''₁(triple₃(a, b, c)) -> foldf₂(triple₃(a, b, nil), c)
f'₂(triple₃(a, b, c), B) -> f₂(triple₃(a, b, c), A)
f₂(t, x) -> f'₂(t, g₁(x))
foldf₂(x, cons₂(y, z)) -> f₂(foldf₂(x, z), y)
f'₂(triple₃(a, b, c), A) -> f''₁(foldf₂(triple₃(cons₂(A, a), nil, c), b))
foldf₂(x, nil) -> x
g₁(A) -> A
g₁(C) -> C
g₁(B) -> A
f'₂(triple₃(a, b, c), C) -> triple₃(a, b, cons₂(C, c))
g₁(B) -> B

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

F₂(t, x) -> F'₂(t, g₁(x))
F'₂(triple₃(a, b, c), A) -> F''₁(foldf₂(triple₃(cons₂(A, a), nil, c), b))
F'₂(triple₃(a, b, c), A) -> FOLDF₂(triple₃(cons₂(A, a), nil, c), b)
F'₂(triple₃(a, b, c), B) -> F₂(triple₃(a, b, c), A)
F''₁(triple₃(a, b, c)) -> FOLDF₂(triple₃(a, b, nil), c)

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
foldf₂(x, nil) -> x
foldf₂(x, cons₂(y, z)) -> f₂(foldf₂(x, z), y)
f₂(t, x) -> f'₂(t, g₁(x))
f'₂(triple₃(a, b, c), C) -> triple₃(a, b, cons₂(C, c))
f'₂(triple₃(a, b, c), B) -> f₂(triple₃(a, b, c), A)
f'₂(triple₃(a, b, c), A) -> f''₁(foldf₂(triple₃(cons₂(A, a), nil, c), b))
f''₁(triple₃(a, b, c)) -> foldf₂(triple₃(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 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

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

F₂(t, x) -> F'₂(t, g₁(x))
F'₂(triple₃(a, b, c), B) -> F₂(triple₃(a, b, c), A)

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
foldf₂(x, nil) -> x
foldf₂(x, cons₂(y, z)) -> f₂(foldf₂(x, z), y)
f₂(t, x) -> f'₂(t, g₁(x))
f'₂(triple₃(a, b, c), C) -> triple₃(a, b, cons₂(C, c))
f'₂(triple₃(a, b, c), B) -> f₂(triple₃(a, b, c), A)
f'₂(triple₃(a, b, c), A) -> f''₁(foldf₂(triple₃(cons₂(A, a), nil, c), b))
f''₁(triple₃(a, b, c)) -> foldf₂(triple₃(a, b, nil), c)

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.

F'₂(triple₃(a, b, c), B) -> F₂(triple₃(a, b, c), A)

The remaining pairs can at least be oriented weakly.

F₂(t, x) -> F'₂(t, g₁(x))

Used ordering: Polynomial interpretation [21]:

POL(A) = 0
POL(B) = 1
POL(C) = 1
POL(F₂(x₁, x₂)) = 1 + x₂
POL(F'₂(x₁, x₂)) = 1 + x₂
POL(g₁(x₁)) = x₁
POL(triple₃(x₁, x₂, x₃)) = 1

The following usable rules [14] were oriented:

g₁(C) -> B
g₁(A) -> A
g₁(C) -> C
g₁(C) -> A
g₁(B) -> A
g₁(B) -> B

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

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

F₂(t, x) -> F'₂(t, g₁(x))

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
foldf₂(x, nil) -> x
foldf₂(x, cons₂(y, z)) -> f₂(foldf₂(x, z), y)
f₂(t, x) -> f'₂(t, g₁(x))
f'₂(triple₃(a, b, c), C) -> triple₃(a, b, cons₂(C, c))
f'₂(triple₃(a, b, c), B) -> f₂(triple₃(a, b, c), A)
f'₂(triple₃(a, b, c), A) -> f''₁(foldf₂(triple₃(cons₂(A, a), nil, c), b))
f''₁(triple₃(a, b, c)) -> foldf₂(triple₃(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 0 SCCs with 1 less node.