Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z126)
The rewrite relation of the following TRS is considered.
|
a(b(a(x1))) |
→ |
a(b(b(a(x1)))) |
(1) |
|
b(b(b(x1))) |
→ |
b(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(b(b(a(x1)))) |
(1) |
|
b(b(b(x1))) |
→ |
b(b(x1)) |
(2) |
1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
|
ab(ba(aa(x1))) |
→ |
ab(bb(ba(aa(x1)))) |
(3) |
|
ab(ba(ab(x1))) |
→ |
ab(bb(ba(ab(x1)))) |
(4) |
|
bb(bb(ba(x1))) |
→ |
bb(ba(x1)) |
(5) |
|
bb(bb(bb(x1))) |
→ |
bb(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(x1)))) |
(7) |
|
ab#(ba(aa(x1))) |
→ |
bb#(ba(aa(x1))) |
(8) |
|
ab#(ba(ab(x1))) |
→ |
ab#(bb(ba(ab(x1)))) |
(9) |
|
ab#(ba(ab(x1))) |
→ |
bb#(ba(ab(x1))) |
(10) |
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.