Certification Problem
Input (TPDB TRS_Relative/Relative_05/rt1-5)
The relative rewrite relation R/S is considered where R is the following TRS
|
s(a(x)) |
→ |
s(b(x)) |
(1) |
|
b(b(x)) |
→ |
a(x) |
(2) |
and S is the following TRS.
|
f(s(x),y) |
→ |
f(x,s(y)) |
(3) |
|
s(b(x)) |
→ |
b(s(x)) |
(4) |
|
b(s(x)) |
→ |
s(b(x)) |
(5) |
|
s(a(x)) |
→ |
a(s(x)) |
(6) |
|
a(s(x)) |
→ |
s(a(x)) |
(7) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [s(x1)] |
= |
1 · x1
|
| [a(x1)] |
= |
1 + 1 · x1
|
| [b(x1)] |
= |
1 + 1 · x1
|
| [f(x1, x2)] |
= |
1 · x1 + 1 · x2
|
all of the following rules can be deleted.
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [s(x1)] |
= |
1 · x1
|
| [a(x1)] |
= |
1 + 1 · x1
|
| [b(x1)] |
= |
1 · x1
|
| [f(x1, x2)] |
= |
1 · x1 + 1 · x2
|
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.