Termination w.r.t. Q proof of TRS/secret06matchbox-gen-10.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:

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

f₁(b₂(a, z)) -> z
b₂(y, b₂(a, z)) -> b₂(f₁(c₃(y, y, a)), b₂(f₁(z), a))
f₁(f₁(f₁(c₃(z, x, a)))) -> b₂(f₁(x), 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 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

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

POL( B₂(x₁, x₂) ) = x₂

POL( a ) = 0

POL( F₁(x₁) ) = x₁

POL( c₃(x₁, ..., x₃) ) = x₁ + x₂

POL( b₂(x₁, x₂) ) = x₂ + 1

POL( f₁(x₁) ) = x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

F₁(f₁(f₁(c₃(z, x, a)))) -> F₁(x)

The TRS R consists of the following rules:

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

F₁(f₁(f₁(c₃(z, x, a)))) -> F₁(x)

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

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

POL( F₁(x₁) ) = x₁ + 1

POL( c₃(x₁, ..., x₃) ) = x₂ + x₃

POL( f₁(x₁) ) = x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

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