Certification Problem
Input (TPDB SRS_Standard/Bouchare_06/02)
The rewrite relation of the following TRS is considered.
|
a(b(a(x1))) |
→ |
b(a(x1)) |
(1) |
|
b(b(b(x1))) |
→ |
b(a(b(x1))) |
(2) |
|
a(a(x1)) |
→ |
b(b(b(x1))) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
a(b(a(x1))) |
→ |
a(b(x1)) |
(4) |
|
b(b(b(x1))) |
→ |
b(a(b(x1))) |
(2) |
|
a(a(x1)) |
→ |
b(b(b(x1))) |
(3) |
1.1 Closure Under Flat Contexts
Using the flat contexts
{a(☐), b(☐)}
We obtain the transformed TRS
|
a(b(a(x1))) |
→ |
a(b(x1)) |
(4) |
|
b(b(b(x1))) |
→ |
b(a(b(x1))) |
(2) |
|
a(a(a(x1))) |
→ |
a(b(b(b(x1)))) |
(5) |
|
b(a(a(x1))) |
→ |
b(b(b(b(x1)))) |
(6) |
1.1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
|
ab(ba(aa(x1))) |
→ |
ab(ba(x1)) |
(7) |
|
ab(ba(ab(x1))) |
→ |
ab(bb(x1)) |
(8) |
|
bb(bb(ba(x1))) |
→ |
ba(ab(ba(x1))) |
(9) |
|
bb(bb(bb(x1))) |
→ |
ba(ab(bb(x1))) |
(10) |
|
aa(aa(aa(x1))) |
→ |
ab(bb(bb(ba(x1)))) |
(11) |
|
aa(aa(ab(x1))) |
→ |
ab(bb(bb(bb(x1)))) |
(12) |
|
ba(aa(aa(x1))) |
→ |
bb(bb(bb(ba(x1)))) |
(13) |
|
ba(aa(ab(x1))) |
→ |
bb(bb(bb(bb(x1)))) |
(14) |
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [ab(x1)] |
= |
1 · x1 + 2 |
| [ba(x1)] |
= |
1 · x1
|
| [aa(x1)] |
= |
1 · x1 + 3 |
| [bb(x1)] |
= |
1 · x1 + 1 |
all of the following rules can be deleted.
|
ab(ba(aa(x1))) |
→ |
ab(ba(x1)) |
(7) |
|
ab(ba(ab(x1))) |
→ |
ab(bb(x1)) |
(8) |
|
aa(aa(aa(x1))) |
→ |
ab(bb(bb(ba(x1)))) |
(11) |
|
aa(aa(ab(x1))) |
→ |
ab(bb(bb(bb(x1)))) |
(12) |
|
ba(aa(aa(x1))) |
→ |
bb(bb(bb(ba(x1)))) |
(13) |
|
ba(aa(ab(x1))) |
→ |
bb(bb(bb(bb(x1)))) |
(14) |
1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [bb(x1)] |
= |
1 · x1 + 1 |
| [ba(x1)] |
= |
1 · x1 + 1 |
| [ab(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
bb(bb(ba(x1))) |
→ |
ba(ab(ba(x1))) |
(9) |
|
bb(bb(bb(x1))) |
→ |
ba(ab(bb(x1))) |
(10) |
1.1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.