Certification Problem
Input (TPDB SRS_Standard/Secret_06_SRS/2-matchbox)
The rewrite relation of the following TRS is considered.
c(c(x1)) |
→ |
a(a(a(b(b(b(x1)))))) |
(1) |
b(b(b(a(x1)))) |
→ |
b(b(b(b(b(b(b(b(x1)))))))) |
(2) |
b(b(c(c(x1)))) |
→ |
c(c(c(a(a(a(a(x1))))))) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1
over the arctic semiring over the integers
[b(x1)] |
= |
· x1 +
|
[c(x1)] |
= |
· x1 +
|
[a(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
c(c(x1)) |
→ |
a(a(a(b(b(b(x1)))))) |
(1) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
b#(b(b(a(x1)))) |
→ |
b#(x1) |
(4) |
b#(b(b(a(x1)))) |
→ |
b#(b(x1)) |
(5) |
b#(b(b(a(x1)))) |
→ |
b#(b(b(x1))) |
(6) |
b#(b(b(a(x1)))) |
→ |
b#(b(b(b(x1)))) |
(7) |
b#(b(b(a(x1)))) |
→ |
b#(b(b(b(b(x1))))) |
(8) |
b#(b(b(a(x1)))) |
→ |
b#(b(b(b(b(b(x1)))))) |
(9) |
b#(b(b(a(x1)))) |
→ |
b#(b(b(b(b(b(b(x1))))))) |
(10) |
b#(b(b(a(x1)))) |
→ |
b#(b(b(b(b(b(b(b(x1)))))))) |
(11) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.