Certification Problem
Input (TPDB SRS_Standard/Trafo_06/un06)
The rewrite relation of the following TRS is considered.
|
b(b(x1)) |
→ |
b(a(b(a(a(b(x1)))))) |
(1) |
|
b(b(a(b(x1)))) |
→ |
b(a(b(b(x1)))) |
(2) |
|
b(a(a(a(b(a(a(b(x1)))))))) |
→ |
b(a(a(a(b(b(x1)))))) |
(3) |
|
b(b(a(a(b(x1))))) |
→ |
b(a(a(b(b(x1))))) |
(4) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
b#(b(x1)) |
→ |
b#(a(b(a(a(b(x1)))))) |
(5) |
|
b#(b(x1)) |
→ |
b#(a(a(b(x1)))) |
(6) |
|
b#(b(a(b(x1)))) |
→ |
b#(a(b(b(x1)))) |
(7) |
|
b#(b(a(b(x1)))) |
→ |
b#(b(x1)) |
(8) |
|
b#(a(a(a(b(a(a(b(x1)))))))) |
→ |
b#(a(a(a(b(b(x1)))))) |
(9) |
|
b#(a(a(a(b(a(a(b(x1)))))))) |
→ |
b#(b(x1)) |
(10) |
|
b#(b(a(a(b(x1))))) |
→ |
b#(a(a(b(b(x1))))) |
(11) |
|
b#(b(a(a(b(x1))))) |
→ |
b#(b(x1)) |
(12) |
1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.
-
The
1st
component contains the
pair
|
b#(a(a(a(b(a(a(b(x1)))))))) |
→ |
b#(a(a(a(b(b(x1)))))) |
(9) |
1.1.1 Reduction Pair Processor
Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals
| [b#(x1)] |
= |
+ · x1
|
| [a(x1)] |
= |
+ · x1
|
| [b(x1)] |
= |
+ · x1
|
the
pair
|
b#(a(a(a(b(a(a(b(x1)))))))) |
→ |
b#(a(a(a(b(b(x1)))))) |
(9) |
could be deleted.
1.1.1.1 P is empty
There are no pairs anymore.
-
The
2nd
component contains the
pair
|
b#(b(a(a(b(x1))))) |
→ |
b#(b(x1)) |
(12) |
|
b#(b(a(b(x1)))) |
→ |
b#(b(x1)) |
(8) |
1.1.2 Size-Change Termination
Using size-change termination in combination with
the subterm criterion
one obtains the following initial size-change graphs.
|
b#(b(a(a(b(x1))))) |
→ |
b#(b(x1)) |
(12) |
|
| 1 |
> |
1 |
|
b#(b(a(b(x1)))) |
→ |
b#(b(x1)) |
(8) |
|
| 1 |
> |
1 |
As there is no critical graph in the transitive closure, there are no infinite chains.