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Edit: Thanks very much to everyone for the answers! As a follow-up question, I'm curious if there are Aristotelian methods for finding contradictories to statements with nested quantifiers. As, thanks to the responses, I now understand the way to go about it using predicate logic. :)

Original post:

Recently, I tried to apply the square of opposition to the phrase "All dogs love all men", but I realized it didn't work. I wrote "Some dogs don't love all men" for the contradictory, but I realized that isn't right. The real contradictory would be "Some dogs don't love some men."

Thus I'm now wondering what the rules are for finding contradictories when there is also a quantifier applying to the predicate term. My personal reflections gave me the following conclusions, but I'm not sure if they're generalizable in a fool-proof way or not. Could someone check this for me and give me feedback? (Ps. here, the double ended arrow is not a biconditional, I am just using it to indicate that the statements are each others' contradictories. Don't know if there's a proper symbol for that)

Contradictory Statements with nested quantifiers

Universal affirmative <--> Particular negative: Predicate quantifier switches

All dogs love all men <–> Some dogs don’t love some men

All dogs love some men <–> Some dogs don’t love all men

Universal negative <–> Particular Affirmative: Predicate quantifier stays the same

No dogs love some men <–> Some dogs love some men

No dogs love all men <–> Some dogs love all men

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    From Aristotelian logic there is no need for double quantifiers. Predicate logic falls closer to math than philosophy. Aristotelian logic did not use symbolization & the concept of what a proposition was strict. The concept of proposition was taken more SERIOUSLY. Today almost anything seems to fit as a proposition which would be rejected in the past. Math made that change. Propositions Express things. It matters what the intent of the message IS-- not how you word it. "All dogs are animals that love human beings" seems to be what your double quantifier claim expresses. That is Aristotelian. – Logikal Sep 4 at 13:37
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    To express some dogs do not love all humans would simply be "Some dogs are animals that love human beings" which means at least one dog loves at least one human being. You can make a compound proposition: some dogs are animals that love human beings & some dogs are not animals that love human beings. One thing to notice is that the Aristotelian way is more descriptive in detail than modern propositions people use. Add more descriptive details about the subject and that usually kills any semantic word play modern people try to come up with & claim fault with Aristotelian logic. – Logikal Sep 4 at 13:50
  • @Logikal - thank you! I'm confused/curious about this phrase: "To express some dogs do not love all humans would simply be "Some dogs are animals that love human beings" which means at least one dog loves at least one human being." Is there a difference in the definition of "some" between Aristotelian and modern logic? I was taught that "some are" does NOT necessarily mean "some are not".... Ie. it could possibly mean "ALL are", but not necessarily (ie. we are not certain that all are) However.... – Lily Sep 5 at 11:46
  • (continuation) Recently I was reading a book that was using Aristotelian logic and they said that "some are" entails "some are not". My former logic prof wrote this off as a careless/uninformed error. But is it in fact a difference in classical/modern definitions? – Lily Sep 5 at 11:47
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Long comment

"All dogs love all men"

is : ∀x ∀y ( Dog(x) ∧ Man(y) → Loves(x,y) ).

Thus, its negation will be : ∃x ∃y ( Dog(x) ∧ Man(y) ∧ ¬ Loves(x,y) ), that reads - as you correctly say :

"Some dogs don't love some men."

Similar for "All dogs love some men", which is : ∀x ∃y ( Dog(x) ∧ Man(y) ∧ Loves(x,y) ).

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Rather than saying, "All dogs love all men", it may be better to think of this as, "All dogs are men-lovers". There is a subject term, dogs, and a predicate term, men-lovers, so that "the predicate is either asserted or denied of the subject". The quantifier is applied to the subject term. Think of the predicate term as an adjective representing a class of objects having that characteristic.

The categorical proposition, "All dogs are men-lovers" would be an 'A' categorical proposition or the universal affirmative. The contradictory would be the 'O' proposition or the particular negative: "Some dogs are not men-lovers".

The 'E' proposition is the universal negative: "No dogs are men-lovers." Its contradictory is the 'I' proposition or the particular affirmative: "Some dogs are men-lovers."

See the Wikipedia page on "syllogism" for how these would be written with quantifiers in predicate logic.


Wikipedia contributors. (2019, January 14). Square of opposition. In Wikipedia, The Free Encyclopedia. Retrieved 04:04, September 2, 2019, from https://en.wikipedia.org/w/index.php?title=Square_of_opposition&oldid=878444149

  • Okay, but the re-phrasing "All dogs are men-lovers" doesn't actually fully capture the proposition "All dogs love all men", as important information is left out.... "All dogs are men-lovers" could very well simply men that all dogs love SOME men.... As for writing it in predicate logic, that is indeed the way I would normally approach this. But I was wondering what people like Fred Sommers, who prefer using traditional logic methods, would do... – Lily Sep 2 at 14:08
  • For some context, the following article by Fred Sommers inspired my question: file:///media/fuse/drivefs-821a8b6148a95084239ce577632be51b/root/aSFU/Phil%20110/Distribution.pdf In the first page, we see that him finding contradictories to statements with nested quantifiers is essential to him being able to use the method of distribution to solve categorical syllogisms. Thank you so much for all your answers Frank Hubeny! – Lily Sep 2 at 14:09
  • @Lily The proposition "All dogs love all men" is unlikely true. There are dogs who bark at people who get near their homes. Nor is it a categorical proposition. The "men-lovers" is a class of all those who love men. It is possible that all dogs might also be men-lovers. It is not likely the case that all dogs are lovers-of-all-men. – Frank Hubeny Sep 2 at 18:33
  • thank you. :) Of course, I know that the statement "all dogs love all men" is very unlikely to be true, haha. I was just trying to figure out what the rules of the game are for finding contradictories for statements of that form. :) – Lily Sep 2 at 21:48
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The idea of a square of opposition relates to Aristotelean logic, which is a logic of categorical propositions. These propositions are allowed to be one of only four figures:

A Every S is P
E No S is P
I Some S is P
O Some S is not P

One of the major problems with this representation is that it is very limiting in terms of what it can express. It restricts propositions to only one quantifier each. As you say in a comment, "all dogs love all men" states more than merely that all dogs are man-lovers. To say the former we need two quantifiers. Another simple example is that Aristotelean categorical logic is unable to express the difference between "every boy loves some girl" and "there is some girl every boy loves", nor can it show that the latter entails the former but not vice versa. Again, these propositions require two quantifiers.

One of the advantages of the predicate logic developed by Frege is that we can greatly expand the range of propositions that can be expressed. The catch is that the square of opposition cannot be fully retained. The A-I and E-O pairs remain contradictories, but the contrary, subcontrary and subaltern relationships no longer hold. Since predicate logic is much more powerful than Aristotelean, the best thing is to learn that and use it to form the contradictories by simply placing a negation (¬) in front.

"All dogs love all men" is represented as

(∀x, ∀y) ( (Dog(x) ∧ Man(y)) → Loves(x,y) ) 

where → is material implication.

Its negation is

¬(∀x, ∀y) ( (Dog(x) ∧ Man(y)) → Loves(x,y) )

which is also equivalent to

(∃x, ∃y) ¬( (Dog(x) ∧ Man(y)) → Loves(x,y) )

and also to

(∃x, ∃y)( Dog(x) ∧ Man(y) ∧ ¬Loves(x,y) )

which we might read as "there is some dog that does not love some man".

"All dogs love some men" is

(∀x)(∃y)( Dog(x) → (Man(y) ∧ Loves(x,y) ) )

and its negation is

¬(∀x)(∃y)( Dog(x) → (Man(y) ∧ Loves(x,y) ) )

which is equivalent to

(∃x)(∀y)¬( Dog(x) → (Man(y) ∧ Loves(x,y) ) )

and also to

(∃x)(∀y)( Dog(x) ∧ ¬( Man(y) ∧ Loves(x,y) ) ) 

which we might read as "there is some dog that does not love any man".

"No dogs love any men" is

(∀x, ∀y) ( (Dog(x) ∧ Man(y)) → ¬Loves(x,y) )

and its negation is

¬(∀x, ∀y) ( (Dog(x) ∧ Man(y)) → ¬Loves(x,y) )

which is equivalent to

(∃x, ∃y) ¬( (Dog(x) ∧ Man(y)) → ¬Loves(x,y) )

and also to

(∃x, ∃y)( Dog(x) ∧ Man(y) ∧ Loves(x,y) )

which we might read as "there is some dog that loves some man".

"No dogs love all men" is

(∀x)(∃y) ( (Dog(x) → (Man(y) ∧ ¬Loves(x,y) ) )

and its negation is

¬(∀x)(∃y) ( (Dog(x) → (Man(y) ∧ ¬Loves(x,y) ) )

which is equivalent to

(∃x)(∀y)¬( Dog(x) → (Man(y) ∧ ¬Loves(x,y) ) )

and also to

(∃x)(∀y)( Dog(x) ∧ ¬( Man(y) ∧ ¬Loves(x,y) ) ) 

and

(∃x)(∀y)( Dog(x) ∧ ( Man(y) → Loves(x,y) ) ) 

which we might read as "there is some dog that loves all men".

  • Thank you! Indeed, I am much more familiar with symbolic predicate logic than with Aristotelian logic. I was just curious how people like Fred Sommers, who still advocate for using Aristotelian logic, would go about solving this... I was reading the following article by him: file:///media/fuse/drivefs-821a8b6148a95084239ce577632be51b/root/aSFU/Phil%20110/Distribution.pdf In the first page, we see that him finding contradictories to statements with nested quantifiers, and I was wondering what his methods for doing that are. Since, as you point out, the square of opposition doesn't apply. – Lily Sep 2 at 21:34
  • Sorry just to clarify: your response does solve my initial question that I posted about, which was simply how to find the contradictories in general :) So thanks very much for that. My curiosity about if there are Aristotelian methods is just a follow-up question, if you happen to know the answer :) – Lily Sep 2 at 21:47
  • As far as I know, it is not possible to express multiply quantified propositions using Aristotelean term logic. Term logic maps onto a fragment of classical many-sorted logic - this was demonstrated by John Corcoran back in the 1970s - so we can do the converse and represent Aristotelean term logic within predicate logic. Also, Peter Strawson in chapter 6 of his Introduction to Logical Theory shows that some aspects of Aristotle's logic do not carry over into predicate logic. For myself, I do not understand the continued interest in Aristotle's logic. Predicate logic has superceded it. – Bumble Sep 2 at 23:03
  • thank you! :D Yes, I don't currently understand the continued interest in Aristotelian logic either, but I'm curious to find out why some still advocate for it. Which is why I want to learn more about it. :) – Lily Sep 2 at 23:12
  • The final section ("Revival") of the Wiki article on Term Logic gives two reasons to account for the continued interest. Good reasons, IMO ;-). en.wikipedia.org/wiki/Term_logic – segfault Sep 12 at 16:18

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