Certification Problem
Input (TPDB SRS_Relative/Waldmann_06_relative/r9)
The relative rewrite relation R/S is considered where R is the following TRS
|
b(a(b(x1))) |
→ |
b(a(a(a(b(x1))))) |
(1) |
and S is the following TRS.
|
a(a(a(x1))) |
→ |
b(b(b(b(x1)))) |
(2) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
matrix interpretations of dimension 6 with strict dimension 1 over the integers
| [b(x1)] |
= |
+
|
| 1 |
0 |
0 |
0 |
0 |
1 |
| 0 |
0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
1 |
| 0 |
0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
0 |
|
|
· x1
|
| [a(x1)] |
= |
+
|
| 1 |
0 |
1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
1 |
0 |
| 0 |
0 |
1 |
0 |
0 |
0 |
| 0 |
0 |
1 |
0 |
0 |
1 |
| 0 |
0 |
0 |
1 |
0 |
0 |
|
|
· x1
|
all of the following rules can be deleted.
|
a(a(a(x1))) |
→ |
b(b(b(b(x1)))) |
(2) |
1.1 Bounds
The given TRS is
match-(raise)-bounded by 1.
This is shown by the following automaton.
-
final states:
{0, 1}
-
transitions:
|
b0(0) |
→ |
0 |
|
b0(1) |
→ |
0 |
|
a0(0) |
→ |
1 |
|
a0(1) |
→ |
1 |
|
b1(0) |
→ |
5 |
|
a1(5) |
→ |
4 |
|
a1(4) |
→ |
3 |
|
a1(3) |
→ |
2 |
|
b1(2) |
→ |
0 |
|
b1(1) |
→ |
5 |
| 0 |
→ |
5 |