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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


B(z, b(c(a, y, a), f(f(x)))) → C(y, a, z)
C(c(z, y, a), a, a) → B(z, y)
B(z, b(c(a, y, a), f(f(x)))) → C(c(y, a, z), z, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
B(x1, x2)  =  B(x1, x2)
b(x1, x2)  =  b(x1, x2)
c(x1, x2, x3)  =  c(x1, x2)
a  =  a
f(x1)  =  x1
C(x1, x2, x3)  =  C(x1, x3)

Recursive Path Order [2].
Precedence:
a > [b2, c2] > [B2, C2]


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP

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

F(c(x, y, z)) → F(b(y, z))

The TRS R consists of the following rules:

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

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