Certification Problem
Input (TPDB SRS_Standard/Trafo_06/un08)
The rewrite relation of the following TRS is considered.
b(b(a(b(x1)))) |
→ |
b(a(b(b(b(b(x1)))))) |
(1) |
b(a(b(b(x1)))) |
→ |
b(b(b(x1))) |
(2) |
b(b(b(x1))) |
→ |
b(b(a(a(b(a(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
b(a(b(b(x1)))) |
→ |
b(b(b(b(a(b(x1)))))) |
(4) |
b(b(a(b(x1)))) |
→ |
b(b(b(x1))) |
(5) |
b(b(b(x1))) |
→ |
b(a(b(a(a(b(b(x1))))))) |
(6) |
1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
ba(ab(bb(bb(x1)))) |
→ |
bb(bb(bb(ba(ab(bb(x1)))))) |
(7) |
ba(ab(bb(ba(x1)))) |
→ |
bb(bb(bb(ba(ab(ba(x1)))))) |
(8) |
bb(ba(ab(bb(x1)))) |
→ |
bb(bb(bb(x1))) |
(9) |
bb(ba(ab(ba(x1)))) |
→ |
bb(bb(ba(x1))) |
(10) |
bb(bb(bb(x1))) |
→ |
ba(ab(ba(aa(ab(bb(bb(x1))))))) |
(11) |
bb(bb(ba(x1))) |
→ |
ba(ab(ba(aa(ab(bb(ba(x1))))))) |
(12) |
1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
ba#(ab(bb(bb(x1)))) |
→ |
bb#(bb(bb(ba(ab(bb(x1)))))) |
(13) |
ba#(ab(bb(bb(x1)))) |
→ |
bb#(bb(ba(ab(bb(x1))))) |
(14) |
ba#(ab(bb(bb(x1)))) |
→ |
bb#(ba(ab(bb(x1)))) |
(15) |
ba#(ab(bb(bb(x1)))) |
→ |
ba#(ab(bb(x1))) |
(16) |
ba#(ab(bb(ba(x1)))) |
→ |
bb#(bb(bb(ba(ab(ba(x1)))))) |
(17) |
ba#(ab(bb(ba(x1)))) |
→ |
bb#(bb(ba(ab(ba(x1))))) |
(18) |
ba#(ab(bb(ba(x1)))) |
→ |
bb#(ba(ab(ba(x1)))) |
(19) |
ba#(ab(bb(ba(x1)))) |
→ |
ba#(ab(ba(x1))) |
(20) |
bb#(ba(ab(bb(x1)))) |
→ |
bb#(bb(bb(x1))) |
(21) |
bb#(ba(ab(bb(x1)))) |
→ |
bb#(bb(x1)) |
(22) |
bb#(ba(ab(ba(x1)))) |
→ |
bb#(bb(ba(x1))) |
(23) |
bb#(ba(ab(ba(x1)))) |
→ |
bb#(ba(x1)) |
(24) |
bb#(bb(bb(x1))) |
→ |
ba#(ab(ba(aa(ab(bb(bb(x1))))))) |
(25) |
bb#(bb(bb(x1))) |
→ |
ba#(aa(ab(bb(bb(x1))))) |
(26) |
bb#(bb(ba(x1))) |
→ |
ba#(ab(ba(aa(ab(bb(ba(x1))))))) |
(27) |
bb#(bb(ba(x1))) |
→ |
ba#(aa(ab(bb(ba(x1))))) |
(28) |
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.