Termination w.r.t. Q proof of TRS/secret06matchbox-gen-15.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:

c₃(z, x, a) -> f₁(b₂(b₂(f₁(z), z), x))
b₂(y, b₂(z, a)) -> f₁(b₂(c₃(f₁(a), y, z), z))
f₁(c₃(c₃(z, a, a), x, a)) -> z

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

c₃(z, x, a) -> f₁(b₂(b₂(f₁(z), z), x))
b₂(y, b₂(z, a)) -> f₁(b₂(c₃(f₁(a), y, z), z))
f₁(c₃(c₃(z, a, a), x, a)) -> z

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:

C₃(z, x, a) -> F₁(b₂(b₂(f₁(z), z), x))
C₃(z, x, a) -> B₂(f₁(z), z)
B₂(y, b₂(z, a)) -> B₂(c₃(f₁(a), y, z), z)
B₂(y, b₂(z, a)) -> C₃(f₁(a), y, z)
C₃(z, x, a) -> B₂(b₂(f₁(z), z), x)
B₂(y, b₂(z, a)) -> F₁(a)
B₂(y, b₂(z, a)) -> F₁(b₂(c₃(f₁(a), y, z), z))
C₃(z, x, a) -> F₁(z)

The TRS R consists of the following rules:

c₃(z, x, a) -> f₁(b₂(b₂(f₁(z), z), x))
b₂(y, b₂(z, a)) -> f₁(b₂(c₃(f₁(a), y, z), z))
f₁(c₃(c₃(z, a, a), x, a)) -> z

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:

C₃(z, x, a) -> F₁(b₂(b₂(f₁(z), z), x))
C₃(z, x, a) -> B₂(f₁(z), z)
B₂(y, b₂(z, a)) -> B₂(c₃(f₁(a), y, z), z)
B₂(y, b₂(z, a)) -> C₃(f₁(a), y, z)
C₃(z, x, a) -> B₂(b₂(f₁(z), z), x)
B₂(y, b₂(z, a)) -> F₁(a)
B₂(y, b₂(z, a)) -> F₁(b₂(c₃(f₁(a), y, z), z))
C₃(z, x, a) -> F₁(z)

The TRS R consists of the following rules:

c₃(z, x, a) -> f₁(b₂(b₂(f₁(z), z), x))
b₂(y, b₂(z, a)) -> f₁(b₂(c₃(f₁(a), y, z), z))
f₁(c₃(c₃(z, a, a), x, a)) -> z

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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

C₃(z, x, a) -> B₂(f₁(z), z)
B₂(y, b₂(z, a)) -> B₂(c₃(f₁(a), y, z), z)
B₂(y, b₂(z, a)) -> C₃(f₁(a), y, z)
C₃(z, x, a) -> B₂(b₂(f₁(z), z), x)

The TRS R consists of the following rules:

c₃(z, x, a) -> f₁(b₂(b₂(f₁(z), z), x))
b₂(y, b₂(z, a)) -> f₁(b₂(c₃(f₁(a), y, z), z))
f₁(c₃(c₃(z, a, a), x, a)) -> z

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.

C₃(z, x, a) -> B₂(f₁(z), z)

The remaining pairs can at least be oriented weakly.

B₂(y, b₂(z, a)) -> B₂(c₃(f₁(a), y, z), z)
B₂(y, b₂(z, a)) -> C₃(f₁(a), y, z)
C₃(z, x, a) -> B₂(b₂(f₁(z), z), x)

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

POL( B₂(x₁, x₂) ) = max{0, 2x₁ + x₂ - 2}

POL( a ) = 3

POL( C₃(x₁, ..., x₃) ) = max{0, 3x₁ + x₂ + x₃ - 1}

POL( c₃(x₁, ..., x₃) ) = 2x₁ + x₂ + 1

POL( f₁(x₁) ) = max{0, x₁ - 3}

POL( b₂(x₁, x₂) ) = x₁ + 2

The following usable rules [14] were oriented:

c₃(z, x, a) -> f₁(b₂(b₂(f₁(z), z), x))
f₁(c₃(c₃(z, a, a), x, a)) -> z
b₂(y, b₂(z, a)) -> f₁(b₂(c₃(f₁(a), y, z), z))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

B₂(y, b₂(z, a)) -> B₂(c₃(f₁(a), y, z), z)
C₃(z, x, a) -> B₂(b₂(f₁(z), z), x)
B₂(y, b₂(z, a)) -> C₃(f₁(a), y, z)

The TRS R consists of the following rules:

c₃(z, x, a) -> f₁(b₂(b₂(f₁(z), z), x))
b₂(y, b₂(z, a)) -> f₁(b₂(c₃(f₁(a), y, z), z))
f₁(c₃(c₃(z, a, a), x, a)) -> z

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.

C₃(z, x, a) -> B₂(b₂(f₁(z), z), x)
B₂(y, b₂(z, a)) -> C₃(f₁(a), y, z)

The remaining pairs can at least be oriented weakly.

B₂(y, b₂(z, a)) -> B₂(c₃(f₁(a), y, z), z)

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

POL( B₂(x₁, x₂) ) = max{0, 2x₁ + 2x₂ - 3}

POL( a ) = 3

POL( C₃(x₁, ..., x₃) ) = max{0, 2x₁ + 2x₂ + 2x₃ - 2}

POL( c₃(x₁, ..., x₃) ) = x₁ + 3

POL( f₁(x₁) ) = max{0, x₁ - 3}

POL( b₂(x₁, x₂) ) = x₁ + 3

The following usable rules [14] were oriented:

c₃(z, x, a) -> f₁(b₂(b₂(f₁(z), z), x))
f₁(c₃(c₃(z, a, a), x, a)) -> z
b₂(y, b₂(z, a)) -> f₁(b₂(c₃(f₁(a), y, z), z))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ QDPOrderProof

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

B₂(y, b₂(z, a)) -> B₂(c₃(f₁(a), y, z), z)

The TRS R consists of the following rules:

c₃(z, x, a) -> f₁(b₂(b₂(f₁(z), z), x))
b₂(y, b₂(z, a)) -> f₁(b₂(c₃(f₁(a), y, z), z))
f₁(c₃(c₃(z, a, a), x, a)) -> z

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.

B₂(y, b₂(z, a)) -> B₂(c₃(f₁(a), y, z), z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( B₂(x₁, x₂) ) = max{0, 3x₂ - 3}

POL( a ) = 3

POL( c₃(x₁, ..., x₃) ) = max{0, 2x₂ + x₃ - 3}

POL( f₁(x₁) ) = x₁ + 2

POL( b₂(x₁, x₂) ) = 3x₁ + 3x₂ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ PisEmptyProof

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

c₃(z, x, a) -> f₁(b₂(b₂(f₁(z), z), x))
b₂(y, b₂(z, a)) -> f₁(b₂(c₃(f₁(a), y, z), z))
f₁(c₃(c₃(z, a, a), x, a)) -> z

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.