Certification Problem
Input (TPDB TRS_Standard/Secret_06_TRS/10)
The rewrite relation of the following TRS is considered.
|
c(c(c(y))) |
→ |
c(c(a(y,0))) |
(1) |
|
c(a(a(0,x),y)) |
→ |
a(c(c(c(0))),y) |
(2) |
|
c(y) |
→ |
y |
(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
| [0] |
= |
|
| [c(x1)] |
= |
· x1 +
|
| [a(x1, x2)] |
= |
· x1 + · x2 +
|
all of the following rules can be deleted.
1.1 Rule Removal
Using the
linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1
over the naturals
| [0] |
= |
|
| 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 |
|
|
|
| [c(x1)] |
= |
|
| 1 |
0 |
1 |
0 |
0 |
| 0 |
0 |
0 |
1 |
0 |
| 0 |
0 |
0 |
0 |
1 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
1 |
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 |
|
|
|
| [a(x1, x2)] |
= |
|
| 1 |
1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
1 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
|
|
· x1 +
|
| 1 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
|
|
· 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 |
|
|
|
all of the following rules can be deleted.
|
c(a(a(0,x),y)) |
→ |
a(c(c(c(0))),y) |
(2) |
1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(a) |
= |
0 |
|
weight(a) |
= |
3 |
|
|
|
| prec(0) |
= |
2 |
|
weight(0) |
= |
1 |
|
|
|
| prec(c) |
= |
3 |
|
weight(c) |
= |
4 |
|
|
|
all of the following rules can be deleted.
|
c(c(c(y))) |
→ |
c(c(a(y,0))) |
(1) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.