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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
          ↳ QDP

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ ATransformationProof
          ↳ QDP

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have applied the A-Transformation [17] to get from an applicative problem to a standard problem. 

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ ATransformationProof
                  ↳ QDP
                    ↳ QDPSizeChangeProof
          ↳ QDP

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

a(b(x)) → a(x)
a(c(x)) → a(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof

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

F(d, f(c, x)) → F(d, f(a, x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ UsableRulesReductionPairsProof

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

F(d, f(c, x)) → F(d, f(a, x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
First, we A-transformed [17] the QDP-Problem. Then we obtain the following A-transformed DP problem.
The pairs P are:

d(c(x)) → d(a(x))

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

a(x) → b(c(x))
a(b(x)) → b(a(x))
a(c(x)) → c(a(x))

Q is empty.

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(a(x1)) = 1 + x1   
POL(b(x1)) = x1   
POL(c(x1)) = 1 + x1   
POL(d(x1)) = 2·x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ UsableRulesReductionPairsProof
                  ↳ QDP
                    ↳ QDPOrderProof

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

d(c(x)) → d(a(x))

The TRS R consists of the following rules:

a(x) → b(c(x))
a(b(x)) → b(a(x))
a(c(x)) → c(a(x))

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.


d(c(x)) → d(a(x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(a(x1)) = x1   
POL(b(x1)) = 0   
POL(c(x1)) = 1 + x1   
POL(d(x1)) = x1   

The following usable rules [17] were oriented:

a(c(x)) → c(a(x))
a(x) → b(c(x))
a(b(x)) → b(a(x))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ UsableRulesReductionPairsProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof

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

a(x) → b(c(x))
a(b(x)) → b(a(x))
a(c(x)) → c(a(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.