Certification Problem
Input (TPDB TRS_Standard/SK90/2.06)
The rewrite relation of the following TRS is considered.
|
+(x,+(y,z)) |
→ |
+(+(x,y),z) |
(1) |
|
+(*(x,y),+(x,z)) |
→ |
*(x,+(y,z)) |
(2) |
|
+(*(x,y),+(*(x,z),u)) |
→ |
+(*(x,+(y,z)),u) |
(3) |
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
| [*(x1, x2)] |
= |
· x1 + · x2 +
|
| [+(x1, x2)] |
= |
· x1 + · x2 +
|
all of the following rules can be deleted.
|
+(*(x,y),+(x,z)) |
→ |
*(x,+(y,z)) |
(2) |
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 +
|
| [+(x1, x2)] |
= |
· x1 + · x2 +
|
all of the following rules can be deleted.
|
+(*(x,y),+(*(x,z),u)) |
→ |
+(*(x,+(y,z)),u) |
(3) |
1.1.1 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 |
0 |
0 |
0 |
| 0 |
0 |
1 |
0 |
| 0 |
0 |
0 |
1 |
|
|
· x1 +
|
| 1 |
0 |
0 |
1 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
1 |
0 |
| 0 |
0 |
0 |
1 |
|
|
· x2 +
|
| 1 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
|
|
|
all of the following rules can be deleted.
|
+(x,+(y,z)) |
→ |
+(+(x,y),z) |
(1) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.