(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, f(b, f(a, x))) → f(a, f(b, f(b, f(a, x))))
f(b, f(b, f(b, x))) → f(b, f(b, 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(b, f(a, x))) → F(a, f(b, f(b, f(a, x))))
F(a, f(b, f(a, x))) → F(b, f(b, f(a, x)))
The TRS R consists of the following rules:
f(a, f(b, f(a, x))) → f(a, f(b, f(b, f(a, x))))
f(b, f(b, f(b, x))) → f(b, f(b, x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(a, f(b, f(a, x))) → F(a, f(b, f(b, f(a, x))))
The TRS R consists of the following rules:
f(a, f(b, f(a, x))) → f(a, f(b, f(b, f(a, x))))
f(b, f(b, f(b, x))) → f(b, f(b, x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) 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(b(a(x))) → a1(b(b(a(x))))
and the Q and R are:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(a(x))) → a(b(b(a(x))))
b(b(b(x))) → b(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.
No dependency pairs are removed.
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [POLO]:
POL(a(x1)) = x1
POL(a1(x1)) = x1
POL(b(x1)) = x1
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
a1(b(a(x))) → a1(b(b(a(x))))
The TRS R consists of the following rules:
a(b(a(x))) → a(b(b(a(x))))
b(b(b(x))) → b(b(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) 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(b(a(x))) → a1(b(b(a(x))))
To find matches we regarded all rules of R and P:
a(b(a(x))) → a(b(b(a(x))))
b(b(b(x))) → b(b(x))
a1(b(a(x))) → a1(b(b(a(x))))
The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
9609785, 9609786, 9609789, 9609788, 9609787, 9609792, 9609790, 9609791
Node 9609785 is start node and node 9609786 is final node.
Those nodes are connect through the following edges:
- 9609785 to 9609787 labelled a1_1(0)
- 9609786 to 9609786 labelled #_1(0)
- 9609789 to 9609786 labelled a_1(0)
- 9609789 to 9609790 labelled a_1(1)
- 9609788 to 9609789 labelled b_1(0)
- 9609787 to 9609788 labelled b_1(0)
- 9609792 to 9609786 labelled a_1(1)
- 9609792 to 9609790 labelled a_1(1)
- 9609790 to 9609791 labelled b_1(1)
- 9609791 to 9609792 labelled b_1(1)
(8) TRUE