Termination w.r.t. Q proof of TRS/secret06matchbox-gen-17.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:

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

f₁(c₃(c₃(a, y, a), b₂(x, z), a)) -> b₂(y, f₁(c₃(f₁(a), z, z)))
f₁(b₂(b₂(x, f₁(y)), z)) -> c₃(z, x, f₁(b₂(b₂(f₁(a), y), y)))
c₃(b₂(a, a), b₂(y, z), x) -> b₂(a, b₂(z, 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 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

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

The TRS R consists of the following rules:

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

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