Certification Problem
Input (TPDB TRS_Standard/SK90/2.37)
The rewrite relation of the following TRS is considered.
|
and(not(not(x)),y,not(z)) |
→ |
and(y,band(x,z),x) |
(1) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1
over the naturals
| [and(x1, x2, x3)] |
= |
|
| 1 |
1 |
0 |
0 |
| 0 |
0 |
0 |
1 |
| 0 |
1 |
0 |
0 |
| 1 |
0 |
0 |
1 |
|
|
· x1 +
|
| 1 |
1 |
0 |
0 |
| 1 |
1 |
0 |
1 |
| 0 |
1 |
0 |
0 |
| 1 |
1 |
1 |
1 |
|
|
· x2 +
|
| 1 |
1 |
0 |
0 |
| 0 |
1 |
0 |
1 |
| 1 |
0 |
0 |
0 |
| 0 |
1 |
0 |
1 |
|
|
· x3 +
|
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
|
|
|
| [not(x1)] |
= |
|
| 1 |
0 |
0 |
0 |
| 1 |
0 |
1 |
1 |
| 0 |
1 |
1 |
1 |
| 1 |
1 |
1 |
0 |
|
|
· x1 +
|
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
|
|
|
| [band(x1, x2)] |
= |
|
| 1 |
0 |
0 |
1 |
| 1 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
| 0 |
0 |
1 |
0 |
|
|
· x1 +
|
| 1 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
1 |
|
|
· x2 +
|
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
|
|
|
all of the following rules can be deleted.
|
and(not(not(x)),y,not(z)) |
→ |
and(y,band(x,z),x) |
(1) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.