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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesReductionPairsProof

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

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

The TRS R consists of the following rules:

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

a1(a(x)) → c1(b(x))
a1(a(x)) → b1(x)
c1(c(x)) → b1(a(x))
b1(b(x)) → c1(x)
b1(b(x)) → a1(c(x))
c1(c(x)) → a1(x)

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

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

The following dependency pairs can be deleted:

a1(a(x)) → b1(x)
b1(b(x)) → c1(x)
c1(c(x)) → a1(x)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(a(x1)) = 2 + 2·x1   
POL(a1(x1)) = 2 + x1   
POL(b(x1)) = 2 + 2·x1   
POL(b1(x1)) = 2 + x1   
POL(c(x1)) = 2 + 2·x1   
POL(c1(x1)) = 2 + x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesReductionPairsProof
        ↳ QDP
          ↳ RFCMatchBoundsDPProof

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

a1(a(x)) → c1(b(x))
c1(c(x)) → b1(a(x))
b1(b(x)) → a1(c(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
Finiteness of the DP problem can be shown by a matchbound of 1.
As the DP problem is minimal we only have to initialize the certificate graph by the rules of P:

a1(a(x)) → c1(b(x))
c1(c(x)) → b1(a(x))
b1(b(x)) → a1(c(x))

To find matches we regarded all rules of R and P:

b(b(x)) → a(c(x))
c(c(x)) → b(a(x))
a(a(x)) → c(b(x))
a1(a(x)) → c1(b(x))
c1(c(x)) → b1(a(x))
b1(b(x)) → a1(c(x))

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

48, 49, 50, 51, 52, 53, 54, 55

Node 48 is start node and node 49 is final node.

Those nodes are connect through the following edges: