Certification Problem
Input (TPDB SRS_Standard/Zantema_06/15)
The rewrite relation of the following TRS is considered.
a(x1) |
→ |
b(x1) |
(1) |
a(a(x1)) |
→ |
a(b(a(x1))) |
(2) |
a(b(x1)) |
→ |
b(b(b(x1))) |
(3) |
a(a(a(x1))) |
→ |
a(a(b(a(a(x1))))) |
(4) |
a(a(b(x1))) |
→ |
a(b(b(a(b(x1))))) |
(5) |
a(b(a(x1))) |
→ |
b(a(b(b(a(x1))))) |
(6) |
a(b(b(x1))) |
→ |
b(b(b(b(b(x1))))) |
(7) |
a(a(a(a(x1)))) |
→ |
a(a(a(b(a(a(a(x1))))))) |
(8) |
a(a(a(b(x1)))) |
→ |
a(a(b(b(a(a(b(x1))))))) |
(9) |
a(a(b(a(x1)))) |
→ |
a(b(a(b(a(b(a(x1))))))) |
(10) |
a(a(b(b(x1)))) |
→ |
a(b(b(b(a(b(b(x1))))))) |
(11) |
a(b(a(a(x1)))) |
→ |
b(a(a(b(b(a(a(x1))))))) |
(12) |
a(b(a(b(x1)))) |
→ |
b(a(b(b(b(a(b(x1))))))) |
(13) |
a(b(b(a(x1)))) |
→ |
b(b(a(b(b(b(a(x1))))))) |
(14) |
a(b(b(b(x1)))) |
→ |
b(b(b(b(b(b(b(x1))))))) |
(15) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by NaTT @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
a#(b(a(x1))) |
→ |
a#(b(b(a(x1)))) |
(16) |
a#(a(b(b(x1)))) |
→ |
a#(b(b(b(a(b(b(x1))))))) |
(17) |
a#(a(a(a(x1)))) |
→ |
a#(a(a(b(a(a(a(x1))))))) |
(18) |
a#(b(b(a(x1)))) |
→ |
a#(b(b(b(a(x1))))) |
(19) |
a#(a(b(a(x1)))) |
→ |
a#(b(a(b(a(x1))))) |
(20) |
a#(a(b(x1))) |
→ |
a#(b(b(a(b(x1))))) |
(21) |
a#(a(a(a(x1)))) |
→ |
a#(a(b(a(a(a(x1)))))) |
(22) |
a#(b(a(a(x1)))) |
→ |
a#(a(b(b(a(a(x1)))))) |
(23) |
a#(a(b(a(x1)))) |
→ |
a#(b(a(b(a(b(a(x1))))))) |
(24) |
a#(a(a(b(x1)))) |
→ |
a#(b(b(a(a(b(x1)))))) |
(25) |
a#(a(a(x1))) |
→ |
a#(b(a(a(x1)))) |
(26) |
a#(a(a(x1))) |
→ |
a#(a(b(a(a(x1))))) |
(27) |
a#(b(a(b(x1)))) |
→ |
a#(b(b(b(a(b(x1)))))) |
(28) |
a#(a(x1)) |
→ |
a#(b(a(x1))) |
(29) |
a#(b(a(a(x1)))) |
→ |
a#(b(b(a(a(x1))))) |
(30) |
a#(a(a(a(x1)))) |
→ |
a#(b(a(a(a(x1))))) |
(31) |
a#(a(a(b(x1)))) |
→ |
a#(a(b(b(a(a(b(x1))))))) |
(32) |
1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.