Certification Problem
Input (TPDB SRS_Standard/Waldmann_19/random-466)
The rewrite relation of the following TRS is considered.
|
a(a(b(a(x1)))) |
→ |
a(b(b(a(x1)))) |
(1) |
|
b(a(b(b(x1)))) |
→ |
b(b(a(b(x1)))) |
(2) |
|
b(a(b(a(x1)))) |
→ |
b(b(a(a(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(a(x1)))) |
→ |
a(b(b(a(x1)))) |
(4) |
|
b(b(a(b(x1)))) |
→ |
b(a(b(b(x1)))) |
(5) |
|
a(b(a(b(x1)))) |
→ |
a(a(b(b(x1)))) |
(6) |
1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
|
ab(ba(aa(aa(x1)))) |
→ |
ab(bb(ba(aa(x1)))) |
(7) |
|
ab(ba(aa(ab(x1)))) |
→ |
ab(bb(ba(ab(x1)))) |
(8) |
|
bb(ba(ab(ba(x1)))) |
→ |
ba(ab(bb(ba(x1)))) |
(9) |
|
bb(ba(ab(bb(x1)))) |
→ |
ba(ab(bb(bb(x1)))) |
(10) |
|
ab(ba(ab(ba(x1)))) |
→ |
aa(ab(bb(ba(x1)))) |
(11) |
|
ab(ba(ab(bb(x1)))) |
→ |
aa(ab(bb(bb(x1)))) |
(12) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [ab(x1)] |
= |
1 · x1
|
| [ba(x1)] |
= |
1 · x1 + 1 |
| [aa(x1)] |
= |
1 · x1 + 1 |
| [bb(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
ab(ba(aa(aa(x1)))) |
→ |
ab(bb(ba(aa(x1)))) |
(7) |
|
ab(ba(aa(ab(x1)))) |
→ |
ab(bb(ba(ab(x1)))) |
(8) |
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [bb(x1)] |
= |
1 · x1
|
| [ba(x1)] |
= |
1 · x1
|
| [ab(x1)] |
= |
1 · x1 + 1 |
| [aa(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
ab(ba(ab(ba(x1)))) |
→ |
aa(ab(bb(ba(x1)))) |
(11) |
|
ab(ba(ab(bb(x1)))) |
→ |
aa(ab(bb(bb(x1)))) |
(12) |
1.1.1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
bb#(ba(ab(ba(x1)))) |
→ |
bb#(ba(x1)) |
(13) |
|
bb#(ba(ab(bb(x1)))) |
→ |
bb#(bb(x1)) |
(14) |
1.1.1.1.1.1 Size-Change Termination
Using size-change termination in combination with
the subterm criterion
one obtains the following initial size-change graphs.
|
bb#(ba(ab(ba(x1)))) |
→ |
bb#(ba(x1)) |
(13) |
|
| 1 |
> |
1 |
|
bb#(ba(ab(bb(x1)))) |
→ |
bb#(bb(x1)) |
(14) |
|
| 1 |
> |
1 |
As there is no critical graph in the transitive closure, there are no infinite chains.