Certification Problem
Input (TPDB TRS_Standard/Secret_06_TRS/5)
The rewrite relation of the following TRS is considered.
|
a(a(y,0),0) |
→ |
y |
(1) |
|
c(c(y)) |
→ |
y |
(2) |
|
c(a(c(c(y)),x)) |
→ |
a(c(c(c(a(x,0)))),y) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1
over the naturals
| [a(x1, x2)] |
= |
|
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
1 |
| 0 |
0 |
1 |
0 |
| 0 |
1 |
0 |
0 |
|
|
· x1 +
|
| 1 |
0 |
0 |
0 |
| 0 |
1 |
0 |
0 |
| 0 |
0 |
1 |
0 |
| 0 |
0 |
0 |
1 |
|
|
· x2 +
|
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
|
|
|
| [0] |
= |
|
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
|
|
|
| [c(x1)] |
= |
|
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
1 |
| 0 |
0 |
1 |
0 |
| 0 |
1 |
0 |
0 |
|
|
· x1 +
|
| 1 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
|
|
|
all of the following rules can be deleted.
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
c#(a(c(c(y)),x)) |
→ |
a#(x,0) |
(4) |
|
c#(a(c(c(y)),x)) |
→ |
c#(a(x,0)) |
(5) |
|
c#(a(c(c(y)),x)) |
→ |
c#(c(a(x,0))) |
(6) |
|
c#(a(c(c(y)),x)) |
→ |
c#(c(c(a(x,0)))) |
(7) |
|
c#(a(c(c(y)),x)) |
→ |
a#(c(c(c(a(x,0)))),y) |
(8) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.