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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

c3(z, x, a) -> f1(b2(b2(f1(z), z), x))
b2(y, b2(z, a)) -> f1(b2(c3(f1(a), y, z), z))
f1(c3(c3(z, a, a), x, a)) -> 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 4 less nodes.

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

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

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

The TRS R consists of the following rules:

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


C3(z, x, a) -> B2(f1(z), z)
B2(y, b2(z, a)) -> B2(c3(f1(a), y, z), z)
C3(z, x, a) -> B2(b2(f1(z), z), x)
The remaining pairs can at least be oriented weakly.

B2(y, b2(z, a)) -> C3(f1(a), y, z)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( C3(x1, ..., x3) ) = 2x1 + x2 + 1


POL( B2(x1, x2) ) = max{0, x1 + x2 - 3}


POL( f1(x1) ) = max{0, x1 - 3}


POL( b2(x1, x2) ) = x1 + x2 + 1


POL( a ) = 3


POL( c3(x1, ..., x3) ) = 2x1 + x2



The following usable rules [14] were oriented:

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



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

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

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

The TRS R consists of the following rules:

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

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.