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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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.


F1(f1(f1(c3(z, x, a)))) -> F1(x)
B2(y, b2(a, z)) -> F1(z)
F1(f1(f1(c3(z, x, a)))) -> B2(f1(x), z)
The remaining pairs can at least be oriented weakly.

B2(y, b2(a, z)) -> B2(f1(c3(y, y, a)), b2(f1(z), a))
Used ordering: Polynomial interpretation [21]:

POL(B2(x1, x2)) = 2 + 2·x2   
POL(F1(x1)) = 1 + 2·x1   
POL(a) = 0   
POL(b2(x1, x2)) = 3 + x2   
POL(c3(x1, x2, x3)) = 2·x1 + 3·x2 + 2·x3   
POL(f1(x1)) = 2 + 3·x1   

The following usable rules [14] were oriented: none



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

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

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

The TRS R consists of the following rules:

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

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