Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

f1(c3(c3(a, y, a), b2(x, z), a)) -> b2(y, f1(c3(f1(a), z, z)))
f1(b2(b2(x, f1(y)), z)) -> c3(z, x, f1(b2(b2(f1(a), y), y)))
c3(b2(a, a), b2(y, z), x) -> b2(a, b2(z, 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 5 less nodes.

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

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

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

The TRS R consists of the following rules:

f1(c3(c3(a, y, a), b2(x, z), a)) -> b2(y, f1(c3(f1(a), z, z)))
f1(b2(b2(x, f1(y)), z)) -> c3(z, x, f1(b2(b2(f1(a), y), y)))
c3(b2(a, a), b2(y, z), x) -> b2(a, b2(z, 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(b2(b2(x, f1(y)), z)) -> F1(b2(b2(f1(a), y), y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

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

The following usable rules [14] were oriented: none



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

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

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

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.