Certification Problem
Input (TPDB TRS_Standard/Rubio_04/nestrec)
The rewrite relation of the following TRS is considered.
|
f(g(X)) |
→ |
g(f(f(X))) |
(1) |
|
f(h(X)) |
→ |
h(g(X)) |
(2) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
g(f(X)) |
→ |
f(f(g(X))) |
(3) |
|
h(f(X)) |
→ |
g(h(X)) |
(4) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [f(x1)] |
= |
· x1 +
|
| [g(x1)] |
= |
· x1 +
|
| [h(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
1.1.1 Rule Removal
Using the
linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1
over the naturals
| [f(x1)] |
= |
|
| 1 |
0 |
1 |
0 |
| 0 |
0 |
0 |
1 |
| 0 |
0 |
0 |
0 |
| 0 |
1 |
1 |
0 |
|
|
· x1 +
|
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
|
|
|
| [g(x1)] |
= |
|
| 1 |
1 |
1 |
1 |
| 0 |
1 |
0 |
1 |
| 0 |
0 |
1 |
0 |
| 0 |
1 |
1 |
1 |
|
|
· x1 +
|
| 0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
|
|
|
all of the following rules can be deleted.
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.