Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 UsableRulesReductionPairsProof (⇔)
↳4 QDP
↳5 RFCMatchBoundsDPProof (⇔)
↳6 TRUE

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

(3) UsableRulesReductionPairsProof (EQUIVALENT transformation)

First, we A-transformed [FROCOS05] 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)
b1(b(x)) → a1(c(x))
b1(b(x)) → c1(x)
c1(c(x)) → b1(a(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 [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] 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 [POLO]:

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

(4) Obligation:

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

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

(5) RFCMatchBoundsDPProof (EQUIVALENT transformation)

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

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

7454411, 7454412, 7454413, 7454414, 7454415, 7454416, 7454417, 7454418

Node 7454411 is start node and node 7454412 is final node.

Those nodes are connect through the following edges:

  • 7454411 to 7454413 labelled c1_1(0)
  • 7454411 to 7454414 labelled b1_1(0)
  • 7454411 to 7454415 labelled a1_1(0)
  • 7454412 to 7454412 labelled #_1(0)
  • 7454413 to 7454412 labelled b_1(0)
  • 7454413 to 7454418 labelled a_1(1)
  • 7454414 to 7454412 labelled a_1(0)
  • 7454414 to 7454417 labelled c_1(1)
  • 7454415 to 7454412 labelled c_1(0)
  • 7454415 to 7454416 labelled b_1(1)
  • 7454416 to 7454412 labelled a_1(1)
  • 7454416 to 7454417 labelled c_1(1)
  • 7454417 to 7454412 labelled b_1(1)
  • 7454417 to 7454418 labelled a_1(1)
  • 7454418 to 7454412 labelled c_1(1)
  • 7454418 to 7454416 labelled b_1(1)

(6) TRUE