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

Q is empty.


↳ QTRS
  ↳ AAECC Innermost

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

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

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

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

The signature Sigma is {f}

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

f(x0, a(b(c(x1))))
f(a(x0), x1)
f(b(x0), x1)
f(c(x0), x1)


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(c(x), y) → F(x, c(y))
F(a(x), y) → F(x, a(y))
F(x, a(b(c(y)))) → F(b(c(a(b(x)))), y)
F(b(x), y) → F(x, b(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, a(b(c(x1))))
f(a(x0), x1)
f(b(x0), x1)
f(c(x0), x1)

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof

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

F(c(x), y) → F(x, c(y))
F(a(x), y) → F(x, a(y))
F(x, a(b(c(y)))) → F(b(c(a(b(x)))), y)
F(b(x), y) → F(x, b(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, a(b(c(x1))))
f(a(x0), x1)
f(b(x0), x1)
f(c(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof

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

F(c(x), y) → F(x, c(y))
F(a(x), y) → F(x, a(y))
F(x, a(b(c(y)))) → F(b(c(a(b(x)))), y)
F(b(x), y) → F(x, b(y))

R is empty.
The set Q consists of the following terms:

f(x0, a(b(c(x1))))
f(a(x0), x1)
f(b(x0), x1)
f(c(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

f(x0, a(b(c(x1))))
f(a(x0), x1)
f(b(x0), x1)
f(c(x0), x1)



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof

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

F(c(x), y) → F(x, c(y))
F(a(x), y) → F(x, a(y))
F(x, a(b(c(y)))) → F(b(c(a(b(x)))), y)
F(b(x), y) → F(x, b(y))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].


↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof

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

a(b(c(x))) → b(a(c(b(x))))

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:

A(b(c(x))) → A(c(b(x)))

The TRS R consists of the following rules:

a(b(c(x))) → b(a(c(b(x))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof

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

A(b(c(x))) → A(c(b(x)))

The TRS R consists of the following rules:

a(b(c(x))) → b(a(c(b(x))))

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.