Certification Problem
Input (TPDB SRS_Standard/Zantema_06/09)
The rewrite relation of the following TRS is considered.
|
a(s(x1)) |
→ |
s(a(x1)) |
(1) |
|
b(a(b(s(x1)))) |
→ |
a(b(s(a(x1)))) |
(2) |
|
b(a(b(b(x1)))) |
→ |
c(s(x1)) |
(3) |
|
c(s(x1)) |
→ |
a(b(a(b(x1)))) |
(4) |
|
a(b(a(a(x1)))) |
→ |
b(a(b(a(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
|
s(a(x1)) |
→ |
a(s(x1)) |
(6) |
|
s(b(a(b(x1)))) |
→ |
a(s(b(a(x1)))) |
(7) |
|
b(b(a(b(x1)))) |
→ |
s(c(x1)) |
(8) |
|
s(c(x1)) |
→ |
b(a(b(a(x1)))) |
(9) |
|
a(a(b(a(x1)))) |
→ |
a(b(a(b(x1)))) |
(10) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [a(x1)] |
= |
2 · x1 + 3 |
| [c(x1)] |
= |
4 · x1 + 11 |
| [s(x1)] |
= |
4 · x1 + 1 |
| [b(x1)] |
= |
2 · x1 + 3 |
all of the following rules can be deleted.
|
s(a(x1)) |
→ |
a(s(x1)) |
(6) |
|
s(b(a(b(x1)))) |
→ |
a(s(b(a(x1)))) |
(7) |
1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
b#(b(a(b(x1)))) |
→ |
s#(c(x1)) |
(11) |
|
s#(c(x1)) |
→ |
a#(x1) |
(12) |
|
s#(c(x1)) |
→ |
b#(a(x1)) |
(13) |
|
s#(c(x1)) |
→ |
a#(b(a(x1))) |
(14) |
|
s#(c(x1)) |
→ |
b#(a(b(a(x1)))) |
(15) |
|
a#(a(b(a(x1)))) |
→ |
b#(x1) |
(16) |
|
a#(a(b(a(x1)))) |
→ |
a#(b(x1)) |
(17) |
|
a#(a(b(a(x1)))) |
→ |
b#(a(b(x1))) |
(18) |
|
a#(a(b(a(x1)))) |
→ |
a#(b(a(b(x1)))) |
(19) |
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.