Certification Problem
Input (TPDB SRS_Standard/Secret_06_SRS/secr6)
The rewrite relation of the following TRS is considered.
|
a(b(c(x1))) |
→ |
c(c(c(b(b(b(a(a(a(x1))))))))) |
(1) |
|
c(b(x1)) |
→ |
a(a(a(x1))) |
(2) |
|
a(x1) |
→ |
x1 |
(3) |
|
b(x1) |
→ |
x1 |
(4) |
|
c(x1) |
→ |
x1 |
(5) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
c(b(a(x1))) |
→ |
a(a(a(b(b(b(c(c(c(x1))))))))) |
(6) |
|
b(c(x1)) |
→ |
a(a(a(x1))) |
(7) |
|
a(x1) |
→ |
x1 |
(3) |
|
b(x1) |
→ |
x1 |
(4) |
|
c(x1) |
→ |
x1 |
(5) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
c#(b(a(x1))) |
→ |
c#(x1) |
(8) |
|
c#(b(a(x1))) |
→ |
c#(c(x1)) |
(9) |
|
c#(b(a(x1))) |
→ |
c#(c(c(x1))) |
(10) |
|
c#(b(a(x1))) |
→ |
b#(c(c(c(x1)))) |
(11) |
|
c#(b(a(x1))) |
→ |
b#(b(c(c(c(x1))))) |
(12) |
|
c#(b(a(x1))) |
→ |
b#(b(b(c(c(c(x1)))))) |
(13) |
|
c#(b(a(x1))) |
→ |
a#(b(b(b(c(c(c(x1))))))) |
(14) |
|
c#(b(a(x1))) |
→ |
a#(a(b(b(b(c(c(c(x1)))))))) |
(15) |
|
c#(b(a(x1))) |
→ |
a#(a(a(b(b(b(c(c(c(x1))))))))) |
(16) |
|
b#(c(x1)) |
→ |
a#(x1) |
(17) |
|
b#(c(x1)) |
→ |
a#(a(x1)) |
(18) |
|
b#(c(x1)) |
→ |
a#(a(a(x1))) |
(19) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.