Certification Problem
Input (TPDB SRS_Relative/Waldmann_19/random-77)
The relative rewrite relation R/S is considered where R is the following TRS
|
b(b(c(x1))) |
→ |
c(a(c(x1))) |
(1) |
|
a(c(b(x1))) |
→ |
b(c(c(x1))) |
(2) |
|
b(b(b(x1))) |
→ |
b(c(a(x1))) |
(3) |
and S is the following TRS.
|
b(a(a(x1))) |
→ |
c(c(c(x1))) |
(4) |
|
a(a(b(x1))) |
→ |
b(a(a(x1))) |
(5) |
|
c(a(b(x1))) |
→ |
b(b(a(x1))) |
(6) |
|
b(a(a(x1))) |
→ |
c(c(b(x1))) |
(7) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [b(x1)] |
= |
2 · x1 + 9 |
| [c(x1)] |
= |
2 · x1 + 7 |
| [a(x1)] |
= |
2 · x1 + 8 |
all of the following rules can be deleted.
|
b(b(c(x1))) |
→ |
c(a(c(x1))) |
(1) |
|
a(c(b(x1))) |
→ |
b(c(c(x1))) |
(2) |
|
b(b(b(x1))) |
→ |
b(c(a(x1))) |
(3) |
|
b(a(a(x1))) |
→ |
c(c(c(x1))) |
(4) |
|
a(a(b(x1))) |
→ |
b(a(a(x1))) |
(5) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.