Certification Problem
Input (TPDB TRS_Relative/Relative_05/rt1-2)
The relative rewrite relation R/S is considered where R is the following TRS
|
f(x) |
→ |
x |
(1) |
|
g(x) |
→ |
x |
(2) |
and S is the following TRS.
|
f(x) |
→ |
g(x) |
(3) |
|
g(x) |
→ |
f(x) |
(4) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1
over the naturals
| [g(x1)] |
= |
|
| 1 |
0 |
0 |
0 |
0 |
| 0 |
1 |
0 |
0 |
0 |
| 0 |
0 |
1 |
0 |
0 |
| 0 |
0 |
0 |
1 |
0 |
| 0 |
0 |
0 |
0 |
1 |
|
|
· x1 +
|
| 1 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
|
|
|
| [f(x1)] |
= |
|
| 1 |
0 |
0 |
0 |
0 |
| 0 |
1 |
0 |
0 |
0 |
| 0 |
0 |
1 |
0 |
0 |
| 0 |
0 |
0 |
1 |
0 |
| 0 |
0 |
0 |
0 |
1 |
|
|
· x1 +
|
| 1 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
|
|
|
all of the following rules can be deleted.
|
f(x) |
→ |
x |
(1) |
|
g(x) |
→ |
x |
(2) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.