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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

A1(B(x)) → A1(x)
C1(A(x)) → A1(C(x))
B1(C(x)) → B1(x)
C1(A(x)) → C1(x)
B1(C(x)) → C1(B(x))
A1(B(x)) → B1(A(x))

The TRS R consists of the following rules:

A(B(x)) → B(A(x))
B(C(x)) → C(B(x))
C(A(x)) → A(C(x))

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


The following pairs can be oriented strictly and are deleted.


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

F(b, f(c, x)) → F(b, x)
F(c, f(a, x)) → F(c, x)
Used ordering: Combined order from the following AFS and order.
A1(x1)  =  x1
B(x1)  =  B(x1)
C1(x1)  =  C1
A(x1)  =  x1
C(x1)  =  C
B1(x1)  =  B1

Recursive path order with status [2].
Precedence:
B1 > B1 > C1 > C

Status:
B1: multiset
C1: multiset
C: multiset
B1: [1]

The following usable rules [14] were oriented:

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



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

F(c, f(a, x)) → F(c, x)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

C(A(x)) → C(x)

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


The following pairs can be oriented strictly and are deleted.


F(c, f(a, x)) → F(c, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
C(x1)  =  C(x1)
A(x1)  =  A(x1)

Recursive path order with status [2].
Precedence:
A1 > C1

Status:
C1: multiset
A1: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof

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

F(b, f(c, x)) → F(b, x)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

B(C(x)) → B(x)

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


The following pairs can be oriented strictly and are deleted.


F(b, f(c, x)) → F(b, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
B(x1)  =  B(x1)
C(x1)  =  C(x1)

Recursive path order with status [2].
Precedence:
C1 > B1

Status:
C1: multiset
B1: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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.