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Discrete Math Textbook img 1

Screenshots from Rosen's discrete Math textbook.

Here's how they define logical equivalence involving quantifiers:

Statements involving predicates and quantifiers are logically equivalent if and only if they have the same truth value no matter which predicates are substituted into these statements and which domain of discourse is used for the variables in these propositional functions. We use the notation S ≡ T to indicate that two statements S and T involving predicates and quantifiers are logically equivalent.

And they also give ∀x(P(x) ∧ Q(x)) ≡ ∀xP(x) ∧ ∀xQ(x) as an example of logical equivalence.

This might be silly question but the domain of x on the LHS of the expression must also match the domain of x on the RHS of the expression, correct?

Kindly please help clarify this for me.

Mauro ALLEGRANZA
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Bob Marley
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    You're posting many screenshots and questions about seemingly trivial things from this book. If you're having this much difficulty it would be better to find a tutor who can help you one on one. This site is not intended for that, as it floods the site with low quality content. You're also not even taking the time to accept good answers that people have kindly provided to you. Please learn more about this site in the FAQ. – blargoner May 20 '24 at 21:25
  • @blargoner I unfortunately don't have the ability to get a tutor, hence I'm relying on such forums to ask questions. Also, I apologize I haven't accepted answers for some of the questions (just went and accepted some). I'm new to the site (just joined 2 weeks ago) and some of them I didn't accept is because I wanted to return to those questions later and first go over other topics. Also, I don't think it's very respectful to say someone's questions are "low quality". I'm trying to learn as a student and ask my questions however I can so far I've had good results. – Bob Marley May 20 '24 at 21:50
  • @blargoner if you could recommend me any online tutor, that be great too. – Bob Marley May 20 '24 at 22:45
  • @blargoner and please do show me specifically where in the FAQ it mentions what ur saying – Bob Marley May 20 '24 at 22:49
  • See Setting the domain of discourse in quantification logic – Mauro ALLEGRANZA May 21 '24 at 08:47
  • And see Proving that universal quantification distributes over conjunction – Mauro ALLEGRANZA May 21 '24 at 08:49
  • @MauroALLEGRANZA Thank you so much for the answer! However, the posts talk about things not in the textbook that seem a bit complicated (like c^D). Also, I'm not sure if they really help answer my questions though (it might be wrong though). My main confusion was whether the domain of discourse of the RHS of logical equivalence statement matches that of the LHS. Additionally, I have another main confusion now.... – Bob Marley May 21 '24 at 15:35
  • @MauroALLEGRANZA ...When we say ∀x(P(x) ∧ Q(x)) ≡ ∀xP(x) ∧ ∀xQ(x), and let elements 'a' and 'b' be part of the domain (assuming domains match on LHS & RHS, which I think they should), then does that mean we can have (P(a) ∧ Q(a)) on LHS and P(a) ∧ Q(b) on RHS? This was also my confusion when seeing the logical equivalence exercises in the textbook too. – Bob Marley May 21 '24 at 15:54
  • Also, when I said "(it might be wrong though)" in my 2nd last comment, I meant to say "(I might be wrong though)" – Bob Marley May 21 '24 at 15:55
  • No Rosen proof is different. It uses an unspecified element a and then conclude. The argument used is an informal one. If you are not satisfied by it, you can review the text and use the linked post. – Mauro ALLEGRANZA May 21 '24 at 16:05
  • @MauroALLEGRANZA No I understand that domain of discourse is specific on interpretation, but my question is that should the domain on RHS match the domain on LHS. I understand we can choose any domain, but my question is if the domain on RHS can be (for e.g.) set of all people, and on LHS the domain (for e.g) be set of all clothes? – Bob Marley May 21 '24 at 16:48
  • @MauroALLEGRANZA Also, I'm not sure we're on the same page on my 2nd question, which is for the general logical equivalence ∀x(P(x) ∧ Q(x)) ≡ ∀xP(x) ∧ ∀xQ(x), is this saying that it's possible for P(a) ∧ Q(a) ≡ P(a) ∧ Q(b), where a & b are elements of some domain S (where here I'm going under assumption that domain of LHS must match domain Of RHS) – Bob Marley May 21 '24 at 16:51
  • Intuitively, if every element of the domain is both P and Q, the every element is also P and every element is also Q, and vice versa. That's all. – Mauro ALLEGRANZA May 21 '24 at 18:10

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