Termination w.r.t. Q proof of TRS/secret06matchbox-gen-14.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(f₁(b₂(x, z)), y) -> f₁(f₁(f₁(b₂(z, b₂(y, z)))))
c₃(f₁(f₁(c₃(x, a, z))), a, y) -> b₂(y, f₁(b₂(a, z)))
b₂(b₂(c₃(b₂(a, a), a, z), f₁(a)), y) -> z

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

B₂(f₁(b₂(x, z)), y) -> B₂(z, b₂(y, z))
B₂(f₁(b₂(x, z)), y) -> B₂(y, z)

The TRS R consists of the following rules:

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

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