Certification Problem
Input (TPDB SRS_Standard/Bouchare_06/18)
The rewrite relation of the following TRS is considered.
|
a(b(a(x1))) |
→ |
b(b(a(x1))) |
(1) |
|
b(b(b(x1))) |
→ |
b(a(x1)) |
(2) |
|
b(b(x1)) |
→ |
a(a(a(x1))) |
(3) |
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.
|
a#(b(a(x1))) |
→ |
b#(b(a(x1))) |
(4) |
|
b#(b(b(x1))) |
→ |
b#(a(x1)) |
(5) |
|
b#(b(b(x1))) |
→ |
a#(x1) |
(6) |
|
b#(b(x1)) |
→ |
a#(a(a(x1))) |
(7) |
|
b#(b(x1)) |
→ |
a#(a(x1)) |
(8) |
|
b#(b(x1)) |
→ |
a#(x1) |
(9) |
1.1 Reduction Pair Processor
Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals
| [a#(x1)] |
= |
+ · x1
|
| [b(x1)] |
= |
+ · x1
|
| [a(x1)] |
= |
+ · x1
|
| [b#(x1)] |
= |
+
|
| 0 |
-∞ |
-∞ |
|
-∞ |
-∞ |
-∞ |
|
-∞ |
-∞ |
-∞ |
|
|
· x1
|
the
pair
|
a#(b(a(x1))) |
→ |
b#(b(a(x1))) |
(4) |
could be deleted.
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.