Certification Problem
Input (TPDB TRS_Standard/AProVE_04/IJCAR_1)
The rewrite relation of the following TRS is considered.
|
div(0,y) |
→ |
0 |
(1) |
|
div(x,y) |
→ |
quot(x,y,y) |
(2) |
|
quot(0,s(y),z) |
→ |
0 |
(3) |
|
quot(s(x),s(y),z) |
→ |
quot(x,y,z) |
(4) |
|
quot(x,0,s(z)) |
→ |
s(div(x,s(z))) |
(5) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
div#(x,y) |
→ |
quot#(x,y,y) |
(6) |
|
quot#(s(x),s(y),z) |
→ |
quot#(x,y,z) |
(7) |
|
quot#(x,0,s(z)) |
→ |
div#(x,s(z)) |
(8) |
1.1 Subterm Criterion Processor
We use the projection
and remove the pairs:
|
quot#(s(x),s(y),z) |
→ |
quot#(x,y,z) |
(7) |
1.1.1 Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the arctic semiring over the integers
| [div#(x1, x2)] |
= |
-∞ · x1 + 1 · x2 + 2 |
| [s(x1)] |
= |
-∞ · x1 + 0 |
| [quot#(x1, x2, x3)] |
= |
-∞ · x1 + 0 · x2 +
-∞ · x3 +
-∞ |
| [0] |
= |
2 |
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pair
|
div#(x,y) |
→ |
quot#(x,y,y) |
(6) |
could be deleted.
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.