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
          ↳ UsableRulesReductionPairsProof

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.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F(x1)) = 2·x1   
POL(a) = 0   
POL(b(x1, x2)) = 2·x1 + 2·x2   
POL(f(x1)) = 2·x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesReductionPairsProof
            ↳ QDP
              ↳ RuleRemovalProof

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))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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


Used ordering: POLO with Polynomial interpretation [25]:

POL(F(x1)) = 2·x1   
POL(a) = 0   
POL(b(x1, x2)) = x1 + 2·x2   
POL(f(x1)) = 1 + 2·x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesReductionPairsProof
            ↳ QDP
              ↳ RuleRemovalProof
                ↳ QDP
                  ↳ PisEmptyProof

Q DP problem:
P is empty.
R is empty.
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.