Certification Problem
Input (TPDB TRS_Standard/HirokawaMiddeldorp_04/t012)
The rewrite relation of the following TRS is considered.
|
minus(minus(x)) |
→ |
x |
(1) |
|
minus(+(x,y)) |
→ |
*(minus(minus(minus(x))),minus(minus(minus(y)))) |
(2) |
|
minus(*(x,y)) |
→ |
+(minus(minus(minus(x))),minus(minus(minus(y)))) |
(3) |
|
f(minus(x)) |
→ |
minus(minus(minus(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 (4 x 4)-matrices with strict dimension 1
over the naturals
| [+(x1, x2)] |
= |
|
| 1 |
0 |
0 |
0 |
| 0 |
1 |
1 |
1 |
| 0 |
1 |
1 |
1 |
| 0 |
1 |
1 |
1 |
|
|
· x1 +
|
| 1 |
0 |
0 |
0 |
| 0 |
1 |
1 |
1 |
| 0 |
1 |
1 |
1 |
| 0 |
1 |
1 |
1 |
|
|
· x2 +
|
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
|
|
|
| [f(x1)] |
= |
|
| 1 |
1 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
|
|
· x1 +
|
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
|
|
|
| [minus(x1)] |
= |
|
| 1 |
1 |
1 |
1 |
| 0 |
1 |
0 |
0 |
| 0 |
0 |
1 |
0 |
| 0 |
0 |
0 |
1 |
|
|
· x1 +
|
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
|
|
|
| [*(x1, x2)] |
= |
|
| 1 |
0 |
0 |
0 |
| 0 |
1 |
1 |
1 |
| 0 |
1 |
1 |
1 |
| 0 |
1 |
1 |
1 |
|
|
· x1 +
|
| 1 |
0 |
0 |
0 |
| 0 |
1 |
1 |
1 |
| 0 |
1 |
1 |
1 |
| 0 |
1 |
1 |
1 |
|
|
· x2 +
|
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
|
|
|
all of the following rules can be deleted.
|
minus(*(x,y)) |
→ |
+(minus(minus(minus(x))),minus(minus(minus(y)))) |
(3) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [+(x1, x2)] |
= |
· x1 + · x2 +
|
| [f(x1)] |
= |
· x1 +
|
| [minus(x1)] |
= |
· x1 +
|
| [*(x1, x2)] |
= |
· x1 + · x2 +
|
all of the following rules can be deleted.
|
minus(+(x,y)) |
→ |
*(minus(minus(minus(x))),minus(minus(minus(y)))) |
(2) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [f(x1)] |
= |
· x1 +
|
| [minus(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [f(x1)] |
= |
8 · x1 + 0 |
| [minus(x1)] |
= |
1 · x1 + 1 |
all of the following rules can be deleted.
|
f(minus(x)) |
→ |
minus(minus(minus(f(x)))) |
(4) |
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.