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

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

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.