Certification Problem
Input (TPDB TRS_Standard/SK90/2.54)
The rewrite relation of the following TRS is considered.
|
f(x,a) |
→ |
x |
(1) |
|
f(x,g(y)) |
→ |
f(g(x),y) |
(2) |
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
| [f(x1, x2)] |
= |
|
| 1 |
0 |
0 |
0 |
| 1 |
1 |
0 |
0 |
| 0 |
1 |
1 |
1 |
| 0 |
1 |
0 |
1 |
|
|
· x1 +
|
| 1 |
1 |
0 |
1 |
| 0 |
1 |
1 |
1 |
| 1 |
1 |
0 |
1 |
| 0 |
0 |
1 |
0 |
|
|
· x2 +
|
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
|
|
|
| [a] |
= |
|
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
|
|
|
| [g(x1)] |
= |
|
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
1 |
0 |
| 0 |
1 |
0 |
1 |
|
|
· x1 +
|
| 1 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
|
|
|
all of the following rules can be deleted.
|
f(x,g(y)) |
→ |
f(g(x),y) |
(2) |
1.1 Rule Removal
Using the
Weighted Path Order with the following precedence and status
| prec(f) |
= |
0 |
|
status(f) |
= |
[2, 1] |
|
list-extension(f) |
= |
Lex |
| prec(a) |
= |
0 |
|
status(a) |
= |
[] |
|
list-extension(a) |
= |
Lex |
and the following
Max-polynomial interpretation
| [f(x1, x2)] |
=
|
2 + 1 · x1 + 1 · x2
|
| [a] |
=
|
max(0) |
all of the following rules can be deleted.
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.