0
$\begingroup$

In Categorical logic (AEIO) I and O are subcontraries. They both can not be False at the same time.

What happens after integrating existential import?

My book says:

Two propositions are subcontraries if they cannot be false together. Yet I- and O-propositions are both false when there is no S. If there is no S, then there is no S that is P and there is no S that is not P. It just is nothing at all. So I- and O-propositions are no more subcontraries.

Lee, Siu-Fan. Logic: A Complete Introduction: Teach Yourself (Complete Introductions) (p. 158). John Murray Press. Kindle Edition.

But I and O propositions always have existential import. And this post confirms that subcontraries retain.

$\endgroup$
1
$\begingroup$

In Categorical logic (AEIO) I and O are subcontraries. They both can not be False at the same time.

What happens after integrating existential import?

Seems like you got this just the wrong way around.

With (i.e. after integrating existential import) Existential import (i.e. that the domain cannot be empty), any I and O pair of statements cannot both be false, and hence are subcontraries.

Without existential import (i.e. before integrating existential import), it is possible for a domain to be empty, and in such a world any existential is false, meaning that any I and O statements would be false, and thus without existential import, any I and O pair of statements are not subcontrary.

That website, by the way, that states that for universals you can do away with existential import, but that for existentials, existential imposrt is retained ... is really weird! This would mean that if I gave you a number of statements involving both existentials as well as universals, then given the existential you do have existential import, but for the universal you don't?!? That's nuts: any logic should have a consistent semantics; you can't just choose to have a different semantics for one statements and a different semantics for another. Indeed, I have never seen anything like what this website is claiming you can do.

$\endgroup$
1
$\begingroup$

There is a difference between the Traditional Square of Oppositions and the modern standard translation of so-called Categorical propositions.

According to Aristotle's Logic, the Universal Affirmative ($\text {Aab}$): "a belongs to all b" implies the corresponding Particular ($\text {Iab}$): "a belongs to some b".

With modern symbolization we have:

$\text {(Aab)} \ \ \forall x (b(x) \to a(x))$

and

$\text {(Iab)} \ \ \exists x (b(x) \land a(x))$.

The issue with so-called Existential Import is that $\text {Aab}$ implies $\text {Iab}$ under the condition that there are $\text A$'s.

The plausible explanation is that, according to Aristotle theory of science, terms are universals and they cannot be empty.


The origin of the Square is Aristotle's De Interpretatione (17b.17–26):

A and O are contradictories [they cannot both be true and they cannot both be false]; E and I are contradictories, and A and E are contraries [they cannot both be true but can both be false].

From this, the rest of the Square follows: I and O are subcontraries [they cannot both be false but can both be true], i.e. they cannot both be false, as well as subalternation [a proposition is a subaltern of another iff it must be true if its superaltern is true, and the superaltern must be false if the subaltern is false].

Subcontraries: suppose that I is false. Then its contradictory, E, is true. So E’s contrary, A, is false. An thus A’s contradictory, O, is true. The conclusion is that I and O cannot be both false.

Subalternation: suppose that A is true. Then its contrary E must be false. But then the E’s contradictory, I, must be true. Thus if the A form is true, so must be the corresponding I.

According to modern translation, $\text {Iab}$ is $\exists x (a(x) \land b(x))$ and $\text {Oab}$ is $\exists x (a(x) \land \lnot b(x))$.

If there are no $a$, both are false. Thus, according to modern point of view, two subcontraries can both be false.

$\endgroup$
1
  • $\begingroup$ In "modern standard translation of so-called Categorical propositions." can both I and O be False? $\endgroup$ – Oleg Dats Jan 20 at 13:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.