(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:
9609620, 9609621, 9609622, 9609623, 9609624, 9609625, 9609626, 9609627
Node 9609620 is start node and node 9609621 is final node.
Those nodes are connect through the following edges:
- 9609620 to 9609622 labelled c1_1(0)
- 9609620 to 9609623 labelled b1_1(0)
- 9609620 to 9609624 labelled a1_1(0)
- 9609621 to 9609621 labelled #_1(0)
- 9609622 to 9609621 labelled b_1(0)
- 9609622 to 9609626 labelled a_1(1)
- 9609623 to 9609621 labelled a_1(0)
- 9609623 to 9609627 labelled c_1(1)
- 9609624 to 9609621 labelled c_1(0)
- 9609624 to 9609625 labelled b_1(1)
- 9609625 to 9609621 labelled a_1(1)
- 9609625 to 9609627 labelled c_1(1)
- 9609626 to 9609621 labelled c_1(1)
- 9609626 to 9609625 labelled b_1(1)
- 9609627 to 9609621 labelled b_1(1)
- 9609627 to 9609626 labelled a_1(1)
(6) TRUE