Termination w.r.t. Q proof of TRS/secret06matchbox-gen-18.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(b₂(y, z), c₃(a, a, a)) -> f₁(c₃(z, y, z))
f₁(b₂(b₂(a, z), c₃(a, x, y))) -> z
c₃(y, x, f₁(z)) -> b₂(f₁(b₂(z, x)), z)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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