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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 9 less nodes.

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


B(a, b(c(z, x, y), a)) → C(y, z, a)
The remaining pairs can at least be oriented weakly.

C(f(c(a, y, a)), x, z) → B(y, b(x, a))
Used ordering: Polynomial interpretation [25,35]:

POL(a) = 0   
POL(C(x1, x2, x3)) = (2)x_1 + (4)x_2 + (3)x_3   
POL(f(x1)) = 1   
POL(b(x1, x2)) = x_1 + (2)x_2   
POL(c(x1, x2, x3)) = (2)x_1 + (2)x_2 + x_3   
POL(B(x1, x2)) = 2 + (2)x_2   
The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented: none



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.

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

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

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

The TRS R consists of the following rules:

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

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