Certification Problem
Input (TPDB SRS_Standard/Zantema_06/10)
The rewrite relation of the following TRS is considered.
|
a(d(x1)) |
→ |
d(b(x1)) |
(1) |
|
a(x1) |
→ |
b(b(b(x1))) |
(2) |
|
d(x1) |
→ |
x1 |
(3) |
|
a(x1) |
→ |
x1 |
(4) |
|
b(d(b(x1))) |
→ |
a(d(x1)) |
(5) |
|
b(c(x1)) |
→ |
c(d(d(x1))) |
(6) |
|
a(c(x1)) |
→ |
b(b(c(d(x1)))) |
(7) |
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
| [a(x1)] |
= |
· x1 +
|
| [c(x1)] |
= |
· x1 +
|
| [d(x1)] |
= |
· x1 +
|
| [b(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
a(c(x1)) |
→ |
b(b(c(d(x1)))) |
(7) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
a#(d(x1)) |
→ |
b#(x1) |
(8) |
|
a#(d(x1)) |
→ |
d#(b(x1)) |
(9) |
|
a#(x1) |
→ |
b#(x1) |
(10) |
|
a#(x1) |
→ |
b#(b(x1)) |
(11) |
|
a#(x1) |
→ |
b#(b(b(x1))) |
(12) |
|
b#(d(b(x1))) |
→ |
d#(x1) |
(13) |
|
b#(d(b(x1))) |
→ |
a#(d(x1)) |
(14) |
|
b#(c(x1)) |
→ |
d#(x1) |
(15) |
|
b#(c(x1)) |
→ |
d#(d(x1)) |
(16) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.