Certification Problem
Input (TPDB SRS_Standard/Secret_06_SRS/secr1)
The rewrite relation of the following TRS is considered.
|
a(b(x1)) |
→ |
b(b(b(b(x1)))) |
(1) |
|
b(a(x1)) |
→ |
a(a(a(a(x1)))) |
(2) |
|
a(x1) |
→ |
x1 |
(3) |
|
b(x1) |
→ |
x1 |
(4) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by matchbox @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
b(a(x1)) |
→ |
b(b(b(b(x1)))) |
(5) |
|
a(b(x1)) |
→ |
a(a(a(a(x1)))) |
(6) |
|
a(x1) |
→ |
x1 |
(3) |
|
b(x1) |
→ |
x1 |
(4) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
b#(a(x1)) |
→ |
b#(x1) |
(7) |
|
b#(a(x1)) |
→ |
b#(b(x1)) |
(8) |
|
b#(a(x1)) |
→ |
b#(b(b(x1))) |
(9) |
|
b#(a(x1)) |
→ |
b#(b(b(b(x1)))) |
(10) |
|
a#(b(x1)) |
→ |
a#(x1) |
(11) |
|
a#(b(x1)) |
→ |
a#(a(x1)) |
(12) |
|
a#(b(x1)) |
→ |
a#(a(a(x1))) |
(13) |
|
a#(b(x1)) |
→ |
a#(a(a(a(x1)))) |
(14) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.