Certification Problem
Input (TPDB SRS_Standard/Trafo_06/dup07)
The rewrite relation of the following TRS is considered.
|
a(a(b(b(x1)))) |
→ |
b(b(c(c(a(a(x1)))))) |
(1) |
|
b(b(c(c(x1)))) |
→ |
c(c(b(b(b(b(x1)))))) |
(2) |
|
a(a(c(c(x1)))) |
→ |
c(c(a(a(b(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(b(a(a(x1)))) |
→ |
a(a(c(c(b(b(x1)))))) |
(4) |
|
c(c(b(b(x1)))) |
→ |
b(b(b(b(c(c(x1)))))) |
(5) |
|
c(c(a(a(x1)))) |
→ |
b(b(a(a(c(c(x1)))))) |
(6) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
b#(b(a(a(x1)))) |
→ |
c#(c(b(b(x1)))) |
(7) |
|
b#(b(a(a(x1)))) |
→ |
c#(b(b(x1))) |
(8) |
|
b#(b(a(a(x1)))) |
→ |
b#(b(x1)) |
(9) |
|
b#(b(a(a(x1)))) |
→ |
b#(x1) |
(10) |
|
c#(c(b(b(x1)))) |
→ |
b#(b(b(b(c(c(x1)))))) |
(11) |
|
c#(c(b(b(x1)))) |
→ |
b#(b(b(c(c(x1))))) |
(12) |
|
c#(c(b(b(x1)))) |
→ |
b#(b(c(c(x1)))) |
(13) |
|
c#(c(b(b(x1)))) |
→ |
b#(c(c(x1))) |
(14) |
|
c#(c(b(b(x1)))) |
→ |
c#(c(x1)) |
(15) |
|
c#(c(b(b(x1)))) |
→ |
c#(x1) |
(16) |
|
c#(c(a(a(x1)))) |
→ |
b#(b(a(a(c(c(x1)))))) |
(17) |
|
c#(c(a(a(x1)))) |
→ |
b#(a(a(c(c(x1))))) |
(18) |
|
c#(c(a(a(x1)))) |
→ |
c#(c(x1)) |
(19) |
|
c#(c(a(a(x1)))) |
→ |
c#(x1) |
(20) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.