Certification Problem
Input (TPDB TRS_Standard/Rubio_04/logarquot)
The rewrite relation of the following TRS is considered.
|
min(X,0) |
→ |
X |
(1) |
|
min(s(X),s(Y)) |
→ |
min(X,Y) |
(2) |
|
quot(0,s(Y)) |
→ |
0 |
(3) |
|
quot(s(X),s(Y)) |
→ |
s(quot(min(X,Y),s(Y))) |
(4) |
|
log(s(0)) |
→ |
0 |
(5) |
|
log(s(s(X))) |
→ |
s(log(s(quot(X,s(s(0)))))) |
(6) |
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.
|
min#(s(X),s(Y)) |
→ |
min#(X,Y) |
(7) |
|
quot#(s(X),s(Y)) |
→ |
min#(X,Y) |
(8) |
|
quot#(s(X),s(Y)) |
→ |
quot#(min(X,Y),s(Y)) |
(9) |
|
log#(s(s(X))) |
→ |
quot#(X,s(s(0))) |
(10) |
|
log#(s(s(X))) |
→ |
log#(s(quot(X,s(s(0))))) |
(11) |
1.1 Dependency Graph Processor
The dependency pairs are split into 3
components.
-
The
1st
component contains the
pair
|
log#(s(s(X))) |
→ |
log#(s(quot(X,s(s(0))))) |
(11) |
1.1.1 Subterm Criterion Processor
We use the projection to multisets
| π(log#)
|
= |
{
1, 1, 1
}
|
| π(quot)
|
= |
{
1, 1
}
|
| π(s)
|
= |
{
1, 1, 1
}
|
| π(min)
|
= |
{
1
}
|
to remove the pairs:
|
log#(s(s(X))) |
→ |
log#(s(quot(X,s(s(0))))) |
(11) |
1.1.1.1 P is empty
There are no pairs anymore.
-
The
2nd
component contains the
pair
|
quot#(s(X),s(Y)) |
→ |
quot#(min(X,Y),s(Y)) |
(9) |
1.1.2 Subterm Criterion Processor
We use the projection to multisets
| π(quot#)
|
= |
{
1
}
|
| π(min)
|
= |
{
1
}
|
to remove the pairs:
|
quot#(s(X),s(Y)) |
→ |
quot#(min(X,Y),s(Y)) |
(9) |
1.1.2.1 P is empty
There are no pairs anymore.
-
The
3rd
component contains the
pair
|
min#(s(X),s(Y)) |
→ |
min#(X,Y) |
(7) |
1.1.3 Size-Change Termination
Using size-change termination in combination with
the subterm criterion
one obtains the following initial size-change graphs.
|
min#(s(X),s(Y)) |
→ |
min#(X,Y) |
(7) |
|
|
| 2 |
> |
2 |
| 1 |
> |
1 |
As there is no critical graph in the transitive closure, there are no infinite chains.