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If suppose we know "All A's are B's" then can we say "some B's are A's".

I know that

"All A's are B's" --> "Some B's are not A's"

But can we say

"Some B's are not A's" -->"Some B's are A's"

closed as unclear what you're asking by Keelan, James Kingsbery, jeroenk, Ben, Swami Vishwananda Oct 14 '15 at 11:54

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    The body of your question seems to ask something else than the title. What is it exactly that you're asking? And the line "All A's are B's" --> "Some B's are not A's" is not valid, logically. – Keelan Sep 19 '13 at 19:05
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    All unicorns are animals. But no animals are unicorns. That'll be five cents, please. – Janet Williams Sep 19 '13 at 22:00
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Here is a simple counterexample:

  • All unicorns are horse-like creatures — true, by the definition of a unicorn;

  • Some horse-like creatures are unicorns — false, because there happens not to be any unicorns.

Every case where your syllogism fails, it will be because the class A is in fact the empty set. The empty set is a subset of all sets; but that doesn't mean that every set contains elements which belong to the empty set. In fact, none do.

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    It is arguable though that your first sentence is neither true nor false, since it is meaningless as there are no unicorns. – Keelan Sep 19 '13 at 19:23
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    Not unless you only allow definition by exemplars. Most people would define "a unicorn" as a horse-like creature which grows a long horn from its forehead. The fact that there happen not to be any such creatures does not make the first sentence meaningless; by definition, any creature which is a unicorn, is a horse-like creature. – Niel de Beaudrap Sep 19 '13 at 19:26
  • Is there an equivocation on the form of existence in here? I.e. we have to interpret bullet (1) as "All (hypothetically existing) unicorns..." and bullet (2) as "Some (real world existing) horse-like ..." Since if you explicitly say "Some hypothetically existing horse-like creatures..." then the proposition is true. – Dave Nov 5 '14 at 19:16
  • @Dave: If unicorns are creatures which are very much like horses but have horns, then any unicorn in any (real or hypothetical) world is a horse-like creature --- interpreting universal quantification via material implication, it is also vacuously satisfied in unicorn-free worlds. Whether there are horse-like creatures which are unicorns (i.e. which have horns and other properties of unicorns) depends on the world, but the meaning of the word 'unicorn', and reasoning about them in the abstract, is agnostic as to the presence of exemplars. – Niel de Beaudrap Nov 5 '14 at 20:55
  • @NieldeBeaudrap I read your previous comment as indicating that your second bulleted proposition should be true -- and thus you do not have a counter example, at least under this way of interpreting quatification. – Dave Nov 5 '14 at 21:01
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A historical footnote to Niel's already complete answer.

In Aristotle's syllogistic the (all → some) inference is valid. The argument given for it in the body of the question is not. In modern axiomatizations of Aristotle's assertoric syllogistic (e.g., Corcoran's), the inference is captured as a rule called "a-i conversion." Modern logics don't validate the move because whenever the plurality (≈ extension) corresponding to A is empty, the universal affirmation will vacuously be true, but since there won't be any As, the particular affirmation will be false.

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    (sorry, downvoted you because the answers changed position when I refreshed. I fixed it) – Ryno Sep 19 '13 at 23:37
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The question states:

∀ x . A ( x ) → B ( x ) ⊢ ∃ y . B ( y ) → ¬ A ( y )

This is false.

Then it asks:

∃ x . B ( x ) → ¬ A ( x ) ⊢ ∃ y . B ( y ) → A ( y )

This is false as well.

The title says:

∀ x . A ( x ) → B ( x ) ⊢ ∃ y . B ( y ) → A ( y )

This is false as well.

We cannot derive those conclusions. They are possible, but not necessary.

  • Just in passing, the point about supporting markup has been discussed a bit on meta. Quick things to note might be that only a handful of other stacks support inline Tex processing; and this is basically owing to the fact that its a pretty expensive add-on. – Joseph Weissman Sep 20 '13 at 0:01
  • @JosephWeissman I'm completely ignorant about how this (infrastructure) works, do you mean expensive in computational terms or monetary terms? In computational terms it should only be expensive when a specific post includes it, otherwise it should mean nothing to the processing. Only a few questions here involve logic, so that should not be a huge overhead, I guess. WRT monetary terms, that would suck, but I don't know how it works, as I said. – Trylks Sep 20 '13 at 10:39
  • As I understand it, it's a pretty heavy add-on to every page load, regardless of whether there's actually math/logic markup that needs to be processed... --At any rate, feel free to join in the discussion on meta! – Joseph Weissman Sep 20 '13 at 14:38
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You simple ask if set A is included in set B will set B contains elements of set A. so lets give an example if A = {1,2,3} and B = {1,2,3,4} ,yes A is included in B and yes there are elements from B which are equal to elements from A

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    incorrect answer, as per the other answers on this page. – Ryno Sep 19 '13 at 23:37
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    A={} is included in B={1,2,3,4} but there is no element from B that is equal to an element of A – miracle173 Sep 20 '13 at 5:11
  • if A = Empty set then A is not included in {1,2,3,4} – Flux Sep 20 '13 at 9:45

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