Certification Problem
Input (TPDB SRS_Relative/Zantema_06_relative/rel10)
The relative rewrite relation R/S is considered where R is the following TRS
|
b(p(b(x1))) |
→ |
b(q(b(x1))) |
(1) |
and S is the following TRS.
|
1(p(0(1(0(x1))))) |
→ |
p(x1) |
(2) |
|
q(x1) |
→ |
0(q(0(x1))) |
(3) |
|
q(x1) |
→ |
1(q(1(x1))) |
(4) |
|
q(x1) |
→ |
0(p(0(x1))) |
(5) |
|
q(x1) |
→ |
1(p(1(x1))) |
(6) |
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
| [1(x1)] |
= |
· x1 +
|
| [p(x1)] |
= |
· x1 +
|
| [0(x1)] |
= |
· x1 +
|
| [b(x1)] |
= |
· x1 +
|
| [q(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
b(p(b(x1))) |
→ |
b(q(b(x1))) |
(1) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.