Certification Problem
Input (TPDB SRS_Relative/Mixed_relative_SRS/zr03)
The relative rewrite relation R/S is considered where R is the following TRS
|
a(a(x1)) |
→ |
a(b(a(x1))) |
(1) |
and S is the following TRS.
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
|
aa(aa(x1)) |
→ |
ab(ba(aa(x1))) |
(3) |
|
aa(ab(x1)) |
→ |
ab(ba(ab(x1))) |
(4) |
|
ba(x1) |
→ |
bb(ba(x1)) |
(5) |
|
bb(x1) |
→ |
bb(bb(x1)) |
(6) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [aa(x1)] |
= |
1 + 1 · x1
|
| [ab(x1)] |
= |
1 · x1
|
| [ba(x1)] |
= |
1 · x1
|
| [bb(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
aa(aa(x1)) |
→ |
ab(ba(aa(x1))) |
(3) |
|
aa(ab(x1)) |
→ |
ab(ba(ab(x1))) |
(4) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.