Certification Problem
Input (TPDB SRS_Standard/Bouchare_06/07)
The rewrite relation of the following TRS is considered.
|
b(a(x1)) |
→ |
b(b(x1)) |
(1) |
|
b(a(b(x1))) |
→ |
b(a(a(x1))) |
(2) |
|
a(a(a(x1))) |
→ |
a(b(b(x1))) |
(3) |
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
|
ba(ab(x1)) |
→ |
bb(bb(x1)) |
(4) |
|
ba(aa(x1)) |
→ |
bb(ba(x1)) |
(5) |
|
ba(ab(bb(x1))) |
→ |
ba(aa(ab(x1))) |
(6) |
|
ba(ab(ba(x1))) |
→ |
ba(aa(aa(x1))) |
(7) |
|
aa(aa(ab(x1))) |
→ |
ab(bb(bb(x1))) |
(8) |
|
aa(aa(aa(x1))) |
→ |
ab(bb(ba(x1))) |
(9) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
ba#(aa(x1)) |
→ |
ba#(x1) |
(10) |
|
ba#(ab(bb(x1))) |
→ |
ba#(aa(ab(x1))) |
(11) |
|
ba#(ab(bb(x1))) |
→ |
aa#(ab(x1)) |
(12) |
|
ba#(ab(ba(x1))) |
→ |
ba#(aa(aa(x1))) |
(13) |
|
ba#(ab(ba(x1))) |
→ |
aa#(aa(x1)) |
(14) |
|
ba#(ab(ba(x1))) |
→ |
aa#(x1) |
(15) |
|
aa#(aa(aa(x1))) |
→ |
ba#(x1) |
(16) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.