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:

f(b(a, z)) → z
b(y, b(a, z)) → b(f(c(y, y, a)), b(f(z), a))
f(f(f(c(z, x, a)))) → b(f(x), z)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(b(a, z)) → z
b(y, b(a, z)) → b(f(c(y, y, a)), b(f(z), a))
f(f(f(c(z, x, a)))) → b(f(x), z)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

f(b(a, z)) → z
b(y, b(a, z)) → b(f(c(y, y, a)), b(f(z), a))
f(f(f(c(z, x, a)))) → b(f(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:

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

The TRS R consists of the following rules:

f(b(a, z)) → z
b(y, b(a, z)) → b(f(c(y, y, a)), b(f(z), a))
f(f(f(c(z, x, a)))) → b(f(x), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

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

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

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

The TRS R consists of the following rules:

f(b(a, z)) → z
b(y, b(a, z)) → b(f(c(y, y, a)), b(f(z), a))
f(f(f(c(z, x, a)))) → b(f(x), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(f(f(c(z, x, a)))) → F(x)
F(f(f(c(z, x, a)))) → B(f(x), z)
B(y, b(a, z)) → F(z)
The remaining pairs can at least be oriented weakly.

B(y, b(a, z)) → B(f(c(y, y, a)), b(f(z), a))
Used ordering: Polynomial interpretation [25]:

POL(B(x1, x2)) = 1 + x2   
POL(F(x1)) = 1 + x1   
POL(a) = 0   
POL(b(x1, x2)) = 1 + x2   
POL(c(x1, x2, x3)) = 1 + x1 + x2   
POL(f(x1)) = 1 + x1   

The following usable rules [17] were oriented: none



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

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

B(y, b(a, z)) → B(f(c(y, y, a)), b(f(z), a))

The TRS R consists of the following rules:

f(b(a, z)) → z
b(y, b(a, z)) → b(f(c(y, y, a)), b(f(z), a))
f(f(f(c(z, x, a)))) → b(f(x), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule B(y, b(a, z)) → B(f(c(y, y, a)), b(f(z), a)) we obtained the following new rules:

B(f(c(y_0, y_1, a)), b(a, a)) → B(f(c(f(c(y_0, y_1, a)), f(c(y_0, y_1, a)), a)), b(f(a), a))



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

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

B(f(c(y_0, y_1, a)), b(a, a)) → B(f(c(f(c(y_0, y_1, a)), f(c(y_0, y_1, a)), a)), b(f(a), a))

The TRS R consists of the following rules:

f(b(a, z)) → z
b(y, b(a, z)) → b(f(c(y, y, a)), b(f(z), a))
f(f(f(c(z, x, a)))) → b(f(x), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.