View source for Reverse-vs-inverse

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))))

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))))

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))))

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)))))))

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)))))))

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 (marked *).

: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 [https://pubmed.ncbi.nlm.nih.gov/15892874/ paper] "Relations in
Biomedical Ontologies".