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prove [(∀a ∈ M : A(a)) ⇒ B] ≡ ∃a ∈ M : (A(a) ⇒ B)

I tried doing it by changing the predicate on the LHS to:

¬A(a) ∨ B

but it ultimatley ended up being wrong. I also tried using the contraposition rule by negating both sides but it also didnt work.

So how can I prove that the RHS = LHS? thx.

edit: forgot to mention that M is a non-empty set edit 2: I was told by the prof that to solve it using the follwing rules:

  • ¬(¬A) ≡ A
  • ¬(A ∧ B) ≡ ¬A ∨ ¬B ¬(A ∨ B) ≡ ¬A ∧ ¬B
  • A ⇒ B ≡ ¬(A ∧ ¬B) ≡ ¬A ∨ B
  • A ⇒ B ≡ ¬B ⇒ ¬A
  • A ⇒ B ≡ A ∧ ¬B ⇒ K
  • (A ⇒ B) ∧ (B ⇒ C) ⇒ (A ⇒ C) ≡ T
  • A ⇔ B ≡ (A ⇒ B) ∧ (B ⇒ A)
  • (A ⇒ B) ∧ A ⇒ B ≡ T
  • A ∧ (B ∨ C) ≡ (A ∧ B) ∨ (A ∧ C) A ∨ (B ∧ C) ≡ (A ∨ B) ∧ (A ∨ C)
  • (A ∨ B) ∨ C ≡ A ∨ (B ∨ C) (=: A ∨ B ∨ C) (A ∧ B) ∧ C ≡ A ∧ (B ∧ C) (=: A ∧ B ∧ C)

edit 2: my post was closed because there was a "duplicate" post however, that post is asking about the negation of the universal quantifier (⊢(∀→)↔∃(→)) which does not make it the same problem as my post

saturnscalypso
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  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Oct 08 '23 at 09:21
  • How do we prove if and only if statements? – AlvinL Oct 08 '23 at 09:29
  • If the LHS is false, then $A(a)$ for all $a\in M$ and $\neg B$. Hence, for any $a$, $A(a) \Rightarrow B$ is false. – AlvinL Oct 08 '23 at 09:33
  • When you negate $\forall a,A(a)$, you obtain $\exists a,\neg A(a)$. Does that help? – AlvinL Oct 08 '23 at 09:38
  • Are you supposed to use any specific method to do this? E.g. using equivalence principles? Some kind of formal proof? Or can you just use the semantics of the operators? – Bram28 Oct 08 '23 at 12:46
  • @Bram28 the professor told us to use the laws of mathematical logic – saturnscalypso Oct 08 '23 at 12:56
  • @AlvinL the thing is that if I negate the quantifiers as well as the predicate (because I believe you cannot negate only one or the other) then the LHS is not equal to the RHS – saturnscalypso Oct 08 '23 at 12:58
  • @saturnscalypso Hmmm, that's still a bit unclear to me ... what does your professor mean by 'laws of mathematical logic'? It sounds like you have to use equivalence principles, like you were already trying to do by rewriting the implication to a disjunction, or using contrapositive laws. If so, what laws do you have available involving quantifiers? Can you please add that to your post (so not just as a comment here) – Bram28 Oct 08 '23 at 13:01
  • @bram28 I just added the laws we were shown in the post – saturnscalypso Oct 08 '23 at 13:58
  • Wow, nothing involving quantifiers? Then it clearly cannot be done, period. – Bram28 Oct 08 '23 at 13:59
  • @bram28 this might be a dumb question but if the question asks me to prove something (eg if LHS = RHS) can I simply say that the statement is not true? – saturnscalypso Oct 08 '23 at 13:59
  • If it is not true, you can (and should) certainly point that out ... but you would also need to demonstrate it using a counterexample. Thing is: this equivalence does hold. But you need some laws involving quantifiers to prove it! Can you look some more at your book or notes? You'll probably have some kind of quantifier negation law: $\neg \forall x : \phi \Leftrightarrow \exists x : \neg \phi$ ... but you'll need a couple of others as well. – Bram28 Oct 08 '23 at 14:01
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    all of the inference rules you specified only apply to propositional logic. in order to prove the statement in your post, you need additional rules from first order logic... such as universal elimination and existential introduction. as it is, you cannot prove your formula without intro and elimination rules for quantifiers. – RyRy the Fly Guy Oct 09 '23 at 00:26
  • @bram28 yeah we learnt the negation of quantifiers as well – saturnscalypso Oct 11 '23 at 07:02
  • @bram28 but if i negate (eg) the LHS then the proposition (A(a) ⇒ B) also gets negated and then its no longer equal because the statement becomes ∃a ∈ M : ¬(A(a) ⇒ B) ≡ ∃a ∈ M : (A(a) ⇒ B) – saturnscalypso Oct 11 '23 at 07:04
  • @bram28 (sorry for pinging you so many times) would it be possible for you to show me your working out on proving the equivalence? – saturnscalypso Oct 11 '23 at 07:06
  • Your Edit 2 is wrong. The duplicate is not "asking about the negation of the universal quantifier". – Anne Bauval Oct 11 '23 at 09:14
  • @AnneBauval You are right that the 'duplicate' is not doing a proof involving the negation of a quantifier. However, the 'duplicate' is still inappropriate here, given that the question is about showing the equivalence using equivalence principles, whereas the 'duplicate' asks for a formal proof. Those are distinct methods ... in particular, a formal proof does not readily translate into a series of equivalences, and so the answers of that 'duplicate post are really not helpful to this question. I am reopening the question. – Bram28 Oct 11 '23 at 11:17
  • @saturnscalypso As the Fly Guy and I have pointed out several times, it can not be done using only the principles that you have provided in the post. You need to have principles involving quantifiers! – Bram28 Oct 11 '23 at 11:18

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