Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z024)
The rewrite relation of the following TRS is considered.
|
a(b(a(x1))) |
→ |
a(a(b(b(a(a(x1)))))) |
(1) |
|
b(a(a(b(x1)))) |
→ |
b(a(b(x1))) |
(2) |
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(a(b(b(a(a(x1)))))) |
(1) |
|
b(a(a(b(x1)))) |
→ |
b(a(b(x1))) |
(2) |
1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
|
ab(ba(aa(x1))) |
→ |
aa(ab(bb(ba(aa(aa(x1)))))) |
(3) |
|
ab(ba(ab(x1))) |
→ |
aa(ab(bb(ba(aa(ab(x1)))))) |
(4) |
|
ba(aa(ab(ba(x1)))) |
→ |
ba(ab(ba(x1))) |
(5) |
|
ba(aa(ab(bb(x1)))) |
→ |
ba(ab(bb(x1))) |
(6) |
1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
ab#(ba(aa(x1))) |
→ |
ab#(bb(ba(aa(aa(x1))))) |
(7) |
|
ab#(ba(aa(x1))) |
→ |
ba#(aa(aa(x1))) |
(8) |
|
ab#(ba(ab(x1))) |
→ |
ab#(bb(ba(aa(ab(x1))))) |
(9) |
|
ab#(ba(ab(x1))) |
→ |
ba#(aa(ab(x1))) |
(10) |
|
ba#(aa(ab(ba(x1)))) |
→ |
ba#(ab(ba(x1))) |
(11) |
|
ba#(aa(ab(bb(x1)))) |
→ |
ba#(ab(bb(x1))) |
(12) |
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.
-
The
1st
component contains the
pair
|
ba#(aa(ab(ba(x1)))) |
→ |
ba#(ab(ba(x1))) |
(11) |
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.
|
ba#(aa(ab(ba(x1)))) |
→ |
ba#(ab(ba(x1))) |
(11) |
|
| 1 |
> |
1 |
As there is no critical graph in the transitive closure, there are no infinite chains.