Termination w.r.t. Q proof of TRS/secret06matchbox-gen-22.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:

b(a, b(c(z, x, y), a)) → b(b(z, c(y, z, a)), x)
f(c(a, b(b(z, a), y), x)) → f(c(x, b(z, x), y))
c(f(c(a, y, a)), x, z) → f(b(b(z, z), f(b(y, b(x, a)))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

b(a, b(c(z, x, y), a)) → b(b(z, c(y, z, a)), x)
f(c(a, b(b(z, a), y), x)) → f(c(x, b(z, x), y))
c(f(c(a, y, a)), x, z) → f(b(b(z, z), f(b(y, b(x, a)))))

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:

B(a, b(c(z, x, y), a)) → B(b(z, c(y, z, a)), x)
C(f(c(a, y, a)), x, z) → B(z, z)
C(f(c(a, y, a)), x, z) → B(b(z, z), f(b(y, b(x, a))))
F(c(a, b(b(z, a), y), x)) → F(c(x, b(z, x), y))
C(f(c(a, y, a)), x, z) → F(b(b(z, z), f(b(y, b(x, a)))))
B(a, b(c(z, x, y), a)) → C(y, z, a)
F(c(a, b(b(z, a), y), x)) → B(z, x)
C(f(c(a, y, a)), x, z) → B(x, a)
C(f(c(a, y, a)), x, z) → B(y, b(x, a))
B(a, b(c(z, x, y), a)) → B(z, c(y, z, a))
C(f(c(a, y, a)), x, z) → F(b(y, b(x, a)))
F(c(a, b(b(z, a), y), x)) → C(x, b(z, x), y)

The TRS R consists of the following rules:

b(a, b(c(z, x, y), a)) → b(b(z, c(y, z, a)), x)
f(c(a, b(b(z, a), y), x)) → f(c(x, b(z, x), y))
c(f(c(a, y, a)), x, z) → f(b(b(z, z), f(b(y, b(x, a)))))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

B(a, b(c(z, x, y), a)) → B(b(z, c(y, z, a)), x)
C(f(c(a, y, a)), x, z) → B(z, z)
C(f(c(a, y, a)), x, z) → B(b(z, z), f(b(y, b(x, a))))
F(c(a, b(b(z, a), y), x)) → F(c(x, b(z, x), y))
C(f(c(a, y, a)), x, z) → F(b(b(z, z), f(b(y, b(x, a)))))
B(a, b(c(z, x, y), a)) → C(y, z, a)
F(c(a, b(b(z, a), y), x)) → B(z, x)
C(f(c(a, y, a)), x, z) → B(x, a)
C(f(c(a, y, a)), x, z) → B(y, b(x, a))
B(a, b(c(z, x, y), a)) → B(z, c(y, z, a))
C(f(c(a, y, a)), x, z) → F(b(y, b(x, a)))
F(c(a, b(b(z, a), y), x)) → C(x, b(z, x), y)

The TRS R consists of the following rules:

b(a, b(c(z, x, y), a)) → b(b(z, c(y, z, a)), x)
f(c(a, b(b(z, a), y), x)) → f(c(x, b(z, x), y))
c(f(c(a, y, a)), x, z) → f(b(b(z, z), f(b(y, b(x, a)))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

B(a, b(c(z, x, y), a)) → C(y, z, a)
F(c(a, b(b(z, a), y), x)) → B(z, x)
C(f(c(a, y, a)), x, z) → B(x, a)
C(f(c(a, y, a)), x, z) → B(y, b(x, a))
F(c(a, b(b(z, a), y), x)) → C(x, b(z, x), y)
B(a, b(c(z, x, y), a)) → B(b(z, c(y, z, a)), x)
F(c(a, b(b(z, a), y), x)) → F(c(x, b(z, x), y))
C(f(c(a, y, a)), x, z) → B(b(z, z), f(b(y, b(x, a))))
C(f(c(a, y, a)), x, z) → B(z, z)
C(f(c(a, y, a)), x, z) → F(b(b(z, z), f(b(y, b(x, a)))))
B(a, b(c(z, x, y), a)) → B(z, c(y, z, a))
C(f(c(a, y, a)), x, z) → F(b(y, b(x, a)))

The TRS R consists of the following rules:

b(a, b(c(z, x, y), a)) → b(b(z, c(y, z, a)), x)
f(c(a, b(b(z, a), y), x)) → f(c(x, b(z, x), y))
c(f(c(a, y, a)), x, z) → f(b(b(z, z), f(b(y, b(x, a)))))

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 9 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

B(a, b(c(z, x, y), a)) → C(y, z, a)
C(f(c(a, y, a)), x, z) → B(y, b(x, a))

The TRS R consists of the following rules:

b(a, b(c(z, x, y), a)) → b(b(z, c(y, z, a)), x)
f(c(a, b(b(z, a), y), x)) → f(c(x, b(z, x), y))
c(f(c(a, y, a)), x, z) → f(b(b(z, z), f(b(y, b(x, a)))))

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.

B(a, b(c(z, x, y), a)) → C(y, z, a)

The remaining pairs can at least be oriented weakly.

C(f(c(a, y, a)), x, z) → B(y, b(x, a))

Used ordering: Combined order from the following AFS and order.
B(x1, x2) = x2
a = a
b(x1, x2) = b(x1, x2)
c(x1, x2, x3) = c(x1, x2, x3)
C(x1, x2, x3) = C(x2, x3)
f(x1) = f(x1)

Lexicographic path order with status [19].
Quasi-Precedence:

[b₂, C_2] > a
c₃ > a
f₁ > a

Status:

C₂: [2,1]
a: multiset
c₃: [3,2,1]
b₂: [2,1]
f₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

              ↳ QDP

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

C(f(c(a, y, a)), x, z) → B(y, b(x, a))

The TRS R consists of the following rules:

b(a, b(c(z, x, y), a)) → b(b(z, c(y, z, a)), x)
f(c(a, b(b(z, a), y), x)) → f(c(x, b(z, x), y))
c(f(c(a, y, a)), x, z) → f(b(b(z, z), f(b(y, b(x, a)))))

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

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

F(c(a, b(b(z, a), y), x)) → F(c(x, b(z, x), y))

The TRS R consists of the following rules:

b(a, b(c(z, x, y), a)) → b(b(z, c(y, z, a)), x)
f(c(a, b(b(z, a), y), x)) → f(c(x, b(z, x), y))
c(f(c(a, y, a)), x, z) → f(b(b(z, z), f(b(y, b(x, a)))))

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