Certification Problem
Input (TPDB SRS_Standard/Zantema_06/04)
The rewrite relation of the following TRS is considered.
|
a(a(b(x1))) |
→ |
b(a(x1)) |
(1) |
|
b(a(a(x1))) |
→ |
a(a(a(b(x1)))) |
(2) |
|
a(c(x1)) |
→ |
c(b(x1)) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
b(a(a(x1))) |
→ |
a(b(x1)) |
(4) |
|
a(a(b(x1))) |
→ |
b(a(a(a(x1)))) |
(5) |
|
c(a(x1)) |
→ |
b(c(x1)) |
(6) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
b#(a(a(x1))) |
→ |
b#(x1) |
(7) |
|
b#(a(a(x1))) |
→ |
a#(b(x1)) |
(8) |
|
a#(a(b(x1))) |
→ |
a#(x1) |
(9) |
|
a#(a(b(x1))) |
→ |
a#(a(x1)) |
(10) |
|
a#(a(b(x1))) |
→ |
a#(a(a(x1))) |
(11) |
|
a#(a(b(x1))) |
→ |
b#(a(a(a(x1)))) |
(12) |
|
c#(a(x1)) |
→ |
c#(x1) |
(13) |
|
c#(a(x1)) |
→ |
b#(c(x1)) |
(14) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.
-
The
1st
component contains the
pair
1.1.1.1 Size-Change Termination
Using size-change termination in combination with
the subterm criterion
one obtains the following initial size-change graphs.
|
c#(a(x1)) |
→ |
c#(x1) |
(13) |
|
| 1 |
> |
1 |
As there is no critical graph in the transitive closure, there are no infinite chains.
-
The
2nd
component contains the
pair
|
b#(a(a(x1))) |
→ |
b#(x1) |
(7) |
|
b#(a(a(x1))) |
→ |
a#(b(x1)) |
(8) |
|
a#(a(b(x1))) |
→ |
a#(x1) |
(9) |
|
a#(a(b(x1))) |
→ |
a#(a(x1)) |
(10) |
|
a#(a(b(x1))) |
→ |
a#(a(a(x1))) |
(11) |
|
a#(a(b(x1))) |
→ |
b#(a(a(a(x1)))) |
(12) |
1.1.1.2 Subterm Criterion Processor
We use the projection to multisets
| π(a#)
|
= |
{
1
}
|
| π(b#)
|
= |
{
1, 1
}
|
| π(a)
|
= |
{
1
}
|
| π(b)
|
= |
{
1, 1
}
|
to remove the pairs:
|
a#(a(b(x1))) |
→ |
a#(x1) |
(9) |
|
a#(a(b(x1))) |
→ |
a#(a(x1)) |
(10) |
|
a#(a(b(x1))) |
→ |
a#(a(a(x1))) |
(11) |
1.1.1.2.1 Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the rationals with delta = 1/64
| [a#(x1)] |
= |
3 · x1 + 7/2 |
| [a(x1)] |
= |
1 · x1 + 1 |
| [b#(x1)] |
= |
3/2 · x1 + 2 |
| [b(x1)] |
= |
1/2 · x1 + 0 |
together with the usable
rules
|
b(a(a(x1))) |
→ |
a(b(x1)) |
(4) |
|
a(a(b(x1))) |
→ |
b(a(a(a(x1)))) |
(5) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pairs
|
b#(a(a(x1))) |
→ |
b#(x1) |
(7) |
|
b#(a(a(x1))) |
→ |
a#(b(x1)) |
(8) |
could be deleted.
1.1.1.2.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.