Certification Problem
Input (TPDB TRS_Relative/Mixed_relative_TRS/ijcar2006)
The relative rewrite relation R/S is considered where R is the following TRS
|
f(a,g(y),z) |
→ |
f(a,y,g(y)) |
(1) |
|
f(b,g(y),z) |
→ |
f(a,y,z) |
(2) |
| a |
→ |
b |
(3) |
and S is the following TRS.
|
f(x,y,z) |
→ |
f(x,y,g(z)) |
(4) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [g(x1)] |
= |
· x1 +
|
| [b] |
= |
|
| [a] |
= |
|
| [f(x1, x2, x3)] |
= |
· x1 + · x2 + · x3 +
|
all of the following rules can be deleted.
|
f(a,g(y),z) |
→ |
f(a,y,g(y)) |
(1) |
|
f(b,g(y),z) |
→ |
f(a,y,z) |
(2) |
1.1 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 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
1 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
1 |
0 |
1 |
|
|
· x1 +
|
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
|
|
|
| [b] |
= |
|
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
|
|
|
| [a] |
= |
|
| 1 |
0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
|
|
|
| [f(x1, x2, x3)] |
= |
|
| 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 |
|
|
· 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 |
|
|
· x2 +
|
| 1 |
0 |
1 |
1 |
1 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
|
|
· x3 +
|
| 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.
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.