Certification Problem
Input (TPDB TRS_Standard/SK90/4.44)
The rewrite relation of the following TRS is considered.
|
f(h(x)) |
→ |
f(i(x)) |
(1) |
|
g(i(x)) |
→ |
g(h(x)) |
(2) |
|
h(a) |
→ |
b |
(3) |
|
i(a) |
→ |
b |
(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
| [a] |
= |
|
| [f(x1)] |
= |
· x1 +
|
| [g(x1)] |
= |
· x1 +
|
| [b] |
= |
|
| [h(x1)] |
= |
· x1 +
|
| [i(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
h(a) |
→ |
b |
(3) |
|
i(a) |
→ |
b |
(4) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1
over the naturals
| [f(x1)] |
= |
|
| 1 |
0 |
1 |
0 |
0 |
| 1 |
0 |
0 |
0 |
0 |
| 0 |
0 |
1 |
1 |
1 |
| 0 |
0 |
1 |
1 |
1 |
| 0 |
0 |
1 |
1 |
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 |
|
|
|
| [g(x1)] |
= |
|
| 1 |
1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
1 |
| 1 |
1 |
0 |
0 |
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 |
|
|
|
| [h(x1)] |
= |
|
| 1 |
0 |
0 |
1 |
1 |
| 0 |
1 |
0 |
0 |
0 |
| 0 |
1 |
0 |
0 |
0 |
| 1 |
0 |
0 |
1 |
0 |
| 0 |
0 |
0 |
0 |
0 |
|
|
· x1 +
|
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
|
|
|
| [i(x1)] |
= |
|
| 1 |
0 |
0 |
1 |
1 |
| 0 |
1 |
0 |
0 |
0 |
| 0 |
1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
1 |
0 |
| 1 |
0 |
0 |
0 |
0 |
|
|
· x1 +
|
| 0 |
0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
0 |
|
|
|
all of the following rules can be deleted.
|
f(h(x)) |
→ |
f(i(x)) |
(1) |
|
g(i(x)) |
→ |
g(h(x)) |
(2) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.