Termination w.r.t. Q proof of /tmp/tmpPUtI4J/matchbox-gen-25.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:

b(a, f(b(b(z, y), a))) → z
c(c(z, x, a), a, y) → f(f(c(y, a, f(c(z, y, x)))))
f(f(c(a, y, z))) → b(y, b(z, z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

b(a, f(b(b(z, y), a))) → z
c(c(z, x, a), a, y) → f(f(c(y, a, f(c(z, y, x)))))
f(f(c(a, y, z))) → b(y, b(z, z))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

C(c(z, x, a), a, y) → F(c(z, y, x))
C(c(z, x, a), a, y) → F(c(y, a, f(c(z, y, x))))
C(c(z, x, a), a, y) → F(f(c(y, a, f(c(z, y, x)))))
C(c(z, x, a), a, y) → C(z, y, x)
F(f(c(a, y, z))) → B(y, b(z, z))
F(f(c(a, y, z))) → B(z, z)
C(c(z, x, a), a, y) → C(y, a, f(c(z, y, x)))

The TRS R consists of the following rules:

b(a, f(b(b(z, y), a))) → z
c(c(z, x, a), a, y) → f(f(c(y, a, f(c(z, y, x)))))
f(f(c(a, y, z))) → b(y, b(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:

C(c(z, x, a), a, y) → F(c(z, y, x))
C(c(z, x, a), a, y) → F(c(y, a, f(c(z, y, x))))
C(c(z, x, a), a, y) → F(f(c(y, a, f(c(z, y, x)))))
C(c(z, x, a), a, y) → C(z, y, x)
F(f(c(a, y, z))) → B(y, b(z, z))
F(f(c(a, y, z))) → B(z, z)
C(c(z, x, a), a, y) → C(y, a, f(c(z, y, x)))

The TRS R consists of the following rules:

b(a, f(b(b(z, y), a))) → z
c(c(z, x, a), a, y) → f(f(c(y, a, f(c(z, y, x)))))
f(f(c(a, y, z))) → b(y, b(z, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

C(c(z, x, a), a, y) → C(z, y, x)
C(c(z, x, a), a, y) → C(y, a, f(c(z, y, x)))

The TRS R consists of the following rules:

b(a, f(b(b(z, y), a))) → z
c(c(z, x, a), a, y) → f(f(c(y, a, f(c(z, y, x)))))
f(f(c(a, y, z))) → b(y, b(z, z))

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