Certification Problem
Input (TPDB TRS_Standard/Endrullis_06/direct)
The rewrite relation of the following TRS is considered.
|
h(x,c(y,z)) |
→ |
h(c(s(y),x),z) |
(1) |
|
h(c(s(x),c(s(0),y)),z) |
→ |
h(y,c(s(0),c(x,z))) |
(2) |
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
| [h(x1, x2)] |
= |
· x1 + · x2 +
|
| [0] |
= |
|
| [c(x1, x2)] |
= |
· x1 + · x2 +
|
| [s(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
h(c(s(x),c(s(0),y)),z) |
→ |
h(y,c(s(0),c(x,z))) |
(2) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1
over the naturals
| [h(x1, x2)] |
= |
|
| 1 |
1 |
0 |
0 |
0 |
| 1 |
1 |
0 |
0 |
0 |
| 1 |
0 |
0 |
1 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
1 |
1 |
0 |
0 |
|
|
· x1 +
|
| 1 |
0 |
1 |
1 |
1 |
| 1 |
1 |
1 |
0 |
1 |
| 1 |
0 |
1 |
1 |
1 |
| 0 |
0 |
0 |
0 |
0 |
| 1 |
1 |
1 |
0 |
1 |
|
|
· x2 +
|
| 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 |
|
|
|
| [c(x1, x2)] |
= |
|
| 1 |
0 |
0 |
0 |
0 |
| 0 |
0 |
1 |
0 |
0 |
| 0 |
0 |
0 |
1 |
1 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
1 |
0 |
0 |
0 |
|
|
· x1 +
|
| 1 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
1 |
0 |
| 0 |
1 |
1 |
0 |
1 |
|
|
· x2 +
|
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
|
|
|
| [s(x1)] |
= |
|
| 1 |
0 |
0 |
0 |
1 |
| 1 |
0 |
0 |
0 |
0 |
| 0 |
1 |
0 |
1 |
0 |
| 0 |
0 |
1 |
0 |
0 |
| 1 |
0 |
0 |
0 |
0 |
|
|
· 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 |
|
|
|
all of the following rules can be deleted.
|
h(x,c(y,z)) |
→ |
h(c(s(y),x),z) |
(1) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.