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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

f(c(c(a, y, a), b(x, z), a)) → b(y, f(c(f(a), z, z)))
f(b(b(x, f(y)), z)) → c(z, x, f(b(b(f(a), y), y)))
c(b(a, a), b(y, z), x) → b(a, b(z, 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 5 less nodes.

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

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

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

The TRS R consists of the following rules:

f(c(c(a, y, a), b(x, z), a)) → b(y, f(c(f(a), z, z)))
f(b(b(x, f(y)), z)) → c(z, x, f(b(b(f(a), y), y)))
c(b(a, a), b(y, z), x) → b(a, b(z, 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(b(b(x, f(y)), z)) → F(b(b(f(a), y), y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(f(x1)) = 3 + (2)x_1   
POL(a) = 0   
POL(b(x1, x2)) = (3)x_1 + (4)x_2   
POL(F(x1)) = (4)x_1   
The value of delta used in the strict ordering is 36.
The following usable rules [17] 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:

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