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:

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

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

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