Reverse-vs-inverse - Revision history

Phismith at 15:00, 5 December 2021

2021-12-05T15:00:38Z

← Older revision		Revision as of 15:00, 5 December 2021
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	Note that the same asymmetry existed in the original class-based		Note that the same asymmetry existed in the original class-based
	definitions of the relations as described in the ~~paper "~~[https://pubmed.ncbi.nlm.nih.gov/15892874/ Relations in		definitions of the relations as described in the [https://pubmed.ncbi.nlm.nih.gov/15892874/ paper] "Relations in
	Biomedical Ontologies].		Biomedical Ontologies".

Phismith at 14:59, 5 December 2021

2021-12-05T14:59:54Z

← Older revision		Revision as of 14:59, 5 December 2021
Line 62:		Line 62:

	Note that the same asymmetry existed in the original class-based		Note that the same asymmetry existed in the original class-based
	definitions of the relations as described in the paper "		definitions of the relations as described in the paper "[https://pubmed.ncbi.nlm.nih.gov/15892874/ Relations in
			Biomedical Ontologies].
	[https://~~genomebiology~~.~~biomedcentral~~.~~com~~/~~articles~~/~~10.1186/gb-2005-6-5-r46~~ Relations in
	Biomedical Ontologies]

Phismith at 14:58, 5 December 2021

2021-12-05T14:58:47Z

← Older revision		Revision as of 14:58, 5 December 2021
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	use the expression InverseOf(participates-in-at-all-times).		use the expression InverseOf(participates-in-at-all-times).

	Note that the same asymmetry existed in the original class-based ~~[https://genomebiology.biomedcentral.com/ testing]~~		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		definitions of the relations as described in the paper "
	Biomedical Ontologies]".
			[https://genomebiology.biomedcentral.com/articles/10.1186/gb-2005-6-5-r46 Relations in
			Biomedical Ontologies]

Phismith at 14:58, 5 December 2021

2021-12-05T14:58:02Z

← Older revision		Revision as of 14:58, 5 December 2021
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	use the expression InverseOf(participates-in-at-all-times).		use the expression InverseOf(participates-in-at-all-times).

	Note that the same asymmetry existed in the original class-based		Note that the same asymmetry existed in the original class-based [https://genomebiology.biomedcentral.com/ testing]
	definitions of the relations as described in the paper "[https://genomebiology.biomedcentral.com/articles/10.1186/gb-2005-6-5-r46 Relations in		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]".		Biomedical Ontologies]".

Phismith at 14:56, 5 December 2021

2021-12-05T14:56:06Z

← Older revision		Revision as of 14:56, 5 December 2021
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	where the arrow points, it's p in the former and q in the latter.		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		We could have defined the inverses as well, but then we would land up with
	two more relations.		two more relations (marked *).

	:1. subject-quantified participates-in-at-all-times		:1. subject-quantified participates-in-at-all-times

Phismith at 14:54, 5 December 2021

2021-12-05T14:54:25Z

← Older revision		Revision as of 14:54, 5 December 2021
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	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:
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	::: (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 ~~rhs~~		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)

Phismith at 14:53, 5 December 2021

2021-12-05T14:53:12Z

Phismith at 14:48, 5 December 2021

2021-12-05T14:48:13Z

← Older revision		Revision as of 14:48, 5 December 2021
Line 3:		Line 3:
	Example:		Example:

	First imagine that we don't have the temporal qualification		First imagine that we don't have the temporal qualification, as in

	(forall (p q)		:(forall (p q)
	(iff (participates-in-at-all-times p q)		:: (iff (participates-in-at-all-times p q)
	(forall (t) (participates-in p q t))))		::: (forall (t) (participates-in p q t))))

	We would form the inverse by swapping p and q on the right-hand side		We would form the inverse by swapping p and q on the right-hand side

Phismith: Created page with "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..."

2021-12-05T14:47:42Z

Created page with "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..."

New page

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

(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 rhs

(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) v
(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* 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 the relations as described in the paper "Relations in
Biomedical Ontologies"

← Older revision		Revision as of 14:53, 5 December 2021
Line 12:		Line 12:
	of the iff		of the iff

	(forall (p q)		:(forall (p q)
	(iff (has-participant-at-all-times p q)		:: (iff (has-participant-at-all-times p q)
	(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)

	(forall (p q)		:(forall (p q)
	(iff (has-participant-at-all-times p q)		:: (iff (has-participant-at-all-times p q)
	(forall (t) (has-participant p q t))))		::: (forall (t) (has-participant p q t))))

	To see the difference between reverse and inverse, we'll		To see the difference between reverse and inverse, we'll create the inverse as we did above
	create the inverse as we did above

	Initially:		Initially:

	(forall (p q)		(forall (p q)
	(iff (participates-in-at-all-times p q)		:: (iff (participates-in-at-all-times p q)
	(and (exists (t) (and (participates-in p q t) (exists-at p t)))		:: (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)))))))		::: (forall (t) (if (exists-at p t) (participates-in p q t)))))))

	Swap p and q on the rhs		Swap p and q on the rhs

	(forall (p q)		:(forall (p q)
	(iff (has-participant-at-all-times(inverse) p q)		:: (iff (has-participant-at-all-times(inverse) p q)
	(and (exists (t) (and (participates-in q p t) (exists-at q t)))		::: (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)))))))		:::: (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)
	(iff (has-participant-at-all-times(inverse) p q) v		:: (iff (has-participant-at-all-times(inverse) p q)
	(and (exists (t) (and (has-participant p q t) (exists-at q t)))		::: (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)))))))		:::: (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)		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.		where the arrow points, it's p in the former and q in the latter.

	We could defined the inverses as well but then we would land up with		We could have defined the inverses as well, but then we would land up with
	two more relations.		two more relations.

	1. subject-quantified participates-in-at-all-times		:1. subject-quantified participates-in-at-all-times
	2. object-quantified has-participant-at-all-times (inverse of 1) *		:2. object-quantified has-participant-at-all-times (inverse of 1) *
	3. subject-quantified has-participant-at-all-times (reverse of 1)		:3. subject-quantified has-participant-at-all-times (reverse of 1)
	4. object-quantified participates-in-at-all-times (inverse of 3) *		: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		It was decided this would be even more confusing. So, in OWL, if you
Line 65:		Line 62:

	Note that the same asymmetry existed in the original class-based		Note that the same asymmetry existed in the original class-based
	definitions the relations as described in the paper "Relations in		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"		Biomedical Ontologies]".