Certification Problem
Input (TPDB TRS_Relative/INVY_15/#3.26_rand)
The relative rewrite relation R/S is considered where R is the following TRS
|
f(x) |
→ |
s(x) |
(1) |
|
f(s(s(x))) |
→ |
s(f(f(x))) |
(2) |
and S is the following TRS.
|
rand(x) |
→ |
x |
(3) |
|
rand(x) |
→ |
rand(s(x)) |
(4) |
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
|
f(x) |
→ |
s(x) |
(1) |
|
s(s(f(x))) |
→ |
f(f(s(x))) |
(5) |
|
rand(x) |
→ |
x |
(3) |
|
rand(x) |
→ |
s(rand(x)) |
(6) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [s(x1)] |
= |
· x1 +
|
| [f(x1)] |
= |
· x1 +
|
| [rand(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
1.1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [s(x1)] |
= |
· x1 +
|
| [f(x1)] |
= |
· x1 +
|
| [rand(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
s(s(f(x))) |
→ |
f(f(s(x))) |
(5) |
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1
over the naturals
| [s(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 |
|
|
· 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 |
|
|
|
| [f(x1)] |
= |
|
| 1 |
1 |
0 |
1 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
|
|
· x1 +
|
| 1 |
0 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
|
|
|
| [rand(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 |
|
|
· 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 |
|
|
|
all of the following rules can be deleted.
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.