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 Order [17,21] with Interpretation:

POL( triple₃(x₁, ..., x₃) ) = x₂ + x₃

POL( F''₁(x₁) ) = 3x₁

POL( F₂(x₁, x₂) ) = 3x₁

POL( f''₁(x₁) ) = x₁

POL( F'₂(x₁, x₂) ) = 3x₁

POL( nil ) = 0

POL( g₁(x₁) ) = max{0, -3}

POL( cons₂(x₁, x₂) ) = x₂ + 1

POL( C ) = max{0, -3}

POL( f'₂(x₁, x₂) ) = x₁ + 1

POL( A ) = 0

POL( f₂(x₁, x₂) ) = x₁ + 1

POL( FOLDF₂(x₁, x₂) ) = 3x₁ + 3x₂

POL( foldf₂(x₁, x₂) ) = x₁ + x₂

POL( B ) = 1

The following usable rules [14] were oriented:

f'₂(triple₃(a, b, c), C) -> triple₃(a, b, cons₂(C, c))
foldf₂(x, nil) -> x
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))

↳ 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.

F₂(t, x) -> F'₂(t, g₁(x))

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( triple₃(x₁, ..., x₃) ) = 3x₁ + 3x₂ + 2x₃

POL( A ) = max{0, -1}

POL( F₂(x₁, x₂) ) = 2x₂ + 1

POL( F'₂(x₁, x₂) ) = x₂

POL( g₁(x₁) ) = max{0, 2x₁ - 1}

POL( C ) = 3

POL( B ) = 1

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

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

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 0 SCCs with 1 less node.