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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

B2(a, b2(c3(z, x, y), a)) -> B2(b2(z, c3(y, z, a)), x)
C3(f1(c3(a, y, a)), x, z) -> B2(z, z)
C3(f1(c3(a, y, a)), x, z) -> B2(b2(z, z), f1(b2(y, b2(x, a))))
F1(c3(a, b2(b2(z, a), y), x)) -> F1(c3(x, b2(z, x), y))
C3(f1(c3(a, y, a)), x, z) -> F1(b2(b2(z, z), f1(b2(y, b2(x, a)))))
B2(a, b2(c3(z, x, y), a)) -> C3(y, z, a)
F1(c3(a, b2(b2(z, a), y), x)) -> B2(z, x)
C3(f1(c3(a, y, a)), x, z) -> B2(x, a)
C3(f1(c3(a, y, a)), x, z) -> B2(y, b2(x, a))
B2(a, b2(c3(z, x, y), a)) -> B2(z, c3(y, z, a))
C3(f1(c3(a, y, a)), x, z) -> F1(b2(y, b2(x, a)))
F1(c3(a, b2(b2(z, a), y), x)) -> C3(x, b2(z, x), y)

The TRS R consists of the following rules:

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

B2(a, b2(c3(z, x, y), a)) -> B2(b2(z, c3(y, z, a)), x)
C3(f1(c3(a, y, a)), x, z) -> B2(z, z)
C3(f1(c3(a, y, a)), x, z) -> B2(b2(z, z), f1(b2(y, b2(x, a))))
F1(c3(a, b2(b2(z, a), y), x)) -> F1(c3(x, b2(z, x), y))
C3(f1(c3(a, y, a)), x, z) -> F1(b2(b2(z, z), f1(b2(y, b2(x, a)))))
B2(a, b2(c3(z, x, y), a)) -> C3(y, z, a)
F1(c3(a, b2(b2(z, a), y), x)) -> B2(z, x)
C3(f1(c3(a, y, a)), x, z) -> B2(x, a)
C3(f1(c3(a, y, a)), x, z) -> B2(y, b2(x, a))
B2(a, b2(c3(z, x, y), a)) -> B2(z, c3(y, z, a))
C3(f1(c3(a, y, a)), x, z) -> F1(b2(y, b2(x, a)))
F1(c3(a, b2(b2(z, a), y), x)) -> C3(x, b2(z, x), y)

The TRS R consists of the following rules:

b2(a, b2(c3(z, x, y), a)) -> b2(b2(z, c3(y, z, a)), x)
f1(c3(a, b2(b2(z, a), y), x)) -> f1(c3(x, b2(z, x), y))
c3(f1(c3(a, y, a)), x, z) -> f1(b2(b2(z, z), f1(b2(y, b2(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

B2(a, b2(c3(z, x, y), a)) -> C3(y, z, a)
C3(f1(c3(a, y, a)), x, z) -> B2(y, b2(x, a))

The TRS R consists of the following rules:

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


B2(a, b2(c3(z, x, y), a)) -> C3(y, z, a)
C3(f1(c3(a, y, a)), x, z) -> B2(y, b2(x, a))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(B2(x1, x2)) = 3·x1 + x2   
POL(C3(x1, x2, x3)) = 3 + 3·x1 + 3·x2   
POL(a) = 2   
POL(b2(x1, x2)) = 3·x1 + 2·x2   
POL(c3(x1, x2, x3)) = 1 + 3·x1 + x2 + 3·x3   
POL(f1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



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

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

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

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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP

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

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

The TRS R consists of the following rules:

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

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