Reverse-vs-inverse: Difference between revisions
mNo edit summary |
mNo edit summary |
||
| Line 3: | Line 3: | ||
Example: | Example: | ||
First imagine that we don't have the temporal qualification, as in | First, imagine that we don't have the temporal qualification, as in | ||
:(forall (p q) | :(forall (p q) | ||
| Line 16: | Line 16: | ||
::: (forall (t) (participates-in q p t)))) | ::: (forall (t) (participates-in q p t)))) | ||
Then rewrite participates-in to has-participant, swapping the arguments since | Then rewrite participates-in to has-participant, swapping the arguments, since | ||
(has-participant p q t) =def (participates-in q p t) | (has-participant p q t) =def (participates-in q p t) | ||
| Line 23: | Line 23: | ||
::: (forall (t) (has-participant p q t)))) | ::: (forall (t) (has-participant p q t)))) | ||
To see the difference between reverse and inverse, we'll create the inverse as we did above | To see the difference between reverse and inverse, we'll create the inverse as we did above. | ||
Initially: | Initially: | ||
| Line 32: | Line 32: | ||
::: (forall (t) (if (exists-at p t) (participates-in p q t))))))) | ::: (forall (t) (if (exists-at p t) (participates-in p q t))))))) | ||
Swap p and q on the | Swap p and q on the right-hand side: | ||
:(forall (p q) | :(forall (p q) | ||
| Line 39: | Line 39: | ||
:::: (forall (t) (if (exists-at q t) (participates-in q p t))))))) | :::: (forall (t) (if (exists-at q t) (participates-in q p t))))))) | ||
Now do the participates-in = has-participant swap | Now do the participates-in = has-participant swap: | ||
:(forall (p q) | :(forall (p q) | ||
Revision as of 14:54, 5 December 2021
The temporal scope of an at-all-times relation is the existence of the subject. If we wanted an inverse, the subject becomes the object and and to match the temporal scope it is the existence of subject=now object that qualifies the relation.
Example:
First, imagine that we don't have the temporal qualification, as in
- (forall (p q)
- (iff (participates-in-at-all-times p q)
- (forall (t) (participates-in p q t))))
- (iff (participates-in-at-all-times p q)
We would form the inverse by swapping p and q on the right-hand side of the iff
- (forall (p q)
- (iff (has-participant-at-all-times p q)
- (forall (t) (participates-in q p t))))
- (iff (has-participant-at-all-times p q)
Then rewrite participates-in to has-participant, swapping the arguments, since (has-participant p q t) =def (participates-in q p t)
- (forall (p q)
- (iff (has-participant-at-all-times p q)
- (forall (t) (has-participant p q t))))
- (iff (has-participant-at-all-times p q)
To see the difference between reverse and inverse, we'll create the inverse as we did above.
Initially:
(forall (p q)
- (iff (participates-in-at-all-times p q)
- (and (exists (t) (and (participates-in p q t) (exists-at p t)))
- (forall (t) (if (exists-at p t) (participates-in p q t)))))))
Swap p and q on the right-hand side:
- (forall (p q)
- (iff (has-participant-at-all-times(inverse) p q)
- (and (exists (t) (and (participates-in q p t) (exists-at q t)))
- (forall (t) (if (exists-at q t) (participates-in q p t)))))))
- (and (exists (t) (and (participates-in q p t) (exists-at q t)))
- (iff (has-participant-at-all-times(inverse) p q)
Now do the participates-in = has-participant swap:
- (forall (p q)
- (iff (has-participant-at-all-times(inverse) p q)
- (and (exists (t) (and (has-participant p q t) (exists-at q t)))
- (forall (t) (if (exists-at q t) (has-participant p q t)))))))
- (and (exists (t) (and (has-participant p q t) (exists-at q t)))
- (iff (has-participant-at-all-times(inverse) p q)
If you compare has-participant-at-all-times to has-participant-at-all-times(inverse) where the arrow points, it's p in the former and q in the latter.
We *could* have defined the inverses as well, but then we would land up with two more relations.
- 1. subject-quantified participates-in-at-all-times
- 2. object-quantified has-participant-at-all-times (inverse of 1) *
- 3. subject-quantified has-participant-at-all-times (reverse of 1)
- 4. object-quantified participates-in-at-all-times (inverse of 3) *
It was decided this would be even more confusing. So, in OWL, if you want the inverse of participates-in-at-all-times, which is unnamed, use the expression InverseOf(participates-in-at-all-times).
Note that the same asymmetry existed in the original class-based definitions of the relations as described in the paper "[https://genomebiology.biomedcentral.com/articles/10.1186/gb-2005-6-5-r46 Relations in Biomedical Ontologies]".
