(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:
7454576, 7454577, 7454579, 7454580, 7454578, 7454581, 7454582, 7454583
Node 7454576 is start node and node 7454577 is final node.
Those nodes are connect through the following edges:
- 7454576 to 7454578 labelled a1_1(0)
- 7454577 to 7454577 labelled #_1(0)
- 7454579 to 7454580 labelled b_1(0)
- 7454580 to 7454577 labelled a_1(0)
- 7454580 to 7454581 labelled a_1(1)
- 7454578 to 7454579 labelled b_1(0)
- 7454581 to 7454582 labelled b_1(1)
- 7454582 to 7454583 labelled b_1(1)
- 7454583 to 7454577 labelled a_1(1)
- 7454583 to 7454581 labelled a_1(1)
(8) TRUE