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 [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

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(z, z)
C(f(c(a, y, a)), x, z) → B(b(z, z), f(b(y, b(x, a))))
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)))))
B(a, b(c(z, x, y), a)) → C(y, z, a)
F(c(a, b(b(z, a), y), x)) → B(z, x)
C(f(c(a, y, a)), x, z) → B(x, a)
C(f(c(a, y, a)), x, z) → B(y, b(x, a))
B(a, b(c(z, x, y), a)) → B(z, c(y, z, a))
C(f(c(a, y, a)), x, z) → F(b(y, b(x, a)))
F(c(a, b(b(z, a), y), x)) → 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.
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(a, b(c(z, x, y), a)) → C(y, z, a)
F(c(a, b(b(z, a), y), x)) → B(z, x)
C(f(c(a, y, a)), x, z) → B(x, a)
C(f(c(a, y, a)), x, z) → B(y, b(x, a))
F(c(a, b(b(z, a), y), x)) → C(x, b(z, x), y)
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) → B(b(z, z), f(b(y, b(x, a))))
C(f(c(a, y, a)), x, z) → B(z, z)
C(f(c(a, y, a)), x, z) → F(b(b(z, z), f(b(y, b(x, a)))))
B(a, b(c(z, x, y), a)) → B(z, c(y, z, a))
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 [13,14,18] contains 2 SCCs with 9 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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 [13].


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: Combined order from the following AFS and order.
B(x1, x2)  =  x2
a  =  a
b(x1, x2)  =  x1
c(x1, x2, x3)  =  c(x1, x2, x3)
C(x1, x2, x3)  =  x2
f(x1)  =  x1

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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 [13,14,18] contains 0 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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.