Is A equivalent to B in propositional calculus?

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The discussion revolves around understanding the equivalence of sets A and B in propositional calculus. The original poster seeks clarification on whether their notation correctly expresses that A and B are equivalent sets. They explore the use of "iff" to denote that all elements of A are in B and vice versa, questioning if this implies equivalence or merely subset relationships. A key point raised is the need for clarity on what "equivalent" means, particularly in the context of equality versus other relations like equinumerability. The conversation concludes with the poster affirming their understanding of equivalence under the relation of equality, emphasizing that A and B must share the same elements.
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I am starting to learn propositional calculus and am trying to make sense of the notation. I am trying to express the idea that sets A and B are equivalent. I want to know if the following statement is true and if it shows three equally valid ways of saying that A and B are the same set.

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Thank you for your time. Any help and/or recommendations would be greatly appreciated.

Edit : Looking back at it, I think the first part does not imply that there are no elements of B that are not also in A. It does not eliminate the possibility that A is a subset of B. Should I write :

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Maybe a more direct way would be : ## x \in A ## iff ## x \in B ##.
 
Bacle2 said:
Maybe a more direct way would be : ## x \in A ## iff ## x \in B ##.
Doesn't that only say that all elements of A are also elements of B, making A a subset of B, and not necessarily equivalent to B? Or does using ''iff'' imply that ## x \in B ## iff ## x \in A ## ?
Also, I understand that the way I put it isn't the most direct way of doing it, but I want to know if my usage of these symbols and operators makes sense.

Thank you for your time.
 
If you move the negations inside of
rustynail said:
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you get the axiom of extensionality of Zermelo-Fraenkel. That is, this "iff" is valid.
But it is unclear what you mean by "equivalent". Equivalence requires a relation. Do you mean "equivalent under the relation of equality"? Then that "iff" would be (trivially) valid. But if you mean, say, equinumerability as your equivalence relation, then the implication only goes in one direction. So, what do you mean by "equivalent"?
 
nomadreid said:
If you move the negations inside of

you get the axiom of extensionality of Zermelo-Fraenkel. That is, this "iff" is valid.
But it is unclear what you mean by "equivalent". Equivalence requires a relation. Do you mean "equivalent under the relation of equality"? Then that "iff" would be (trivially) valid. But if you mean, say, equinumerability as your equivalence relation, then the implication only goes in one direction. So, what do you mean by "equ.ivalent"?

I mean ''equivalent under the relation of equality'' as in ''A and B are the same object''. Because A and B share not only the same cardinality, but also the same elements.
So if A = {p, q, r, t}, then B = {p, q, r, t} also, and thus A=B.

Edit : I'm currently looking at the Zermelo-Fraenkel axioms. That's very helpful, thank you!
 
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I'm taking a look at intuitionistic propositional logic (IPL). Basically it exclude Double Negation Elimination (DNE) from the set of axiom schemas replacing it with Ex falso quodlibet: ⊥ → p for any proposition p (including both atomic and composite propositions). In IPL, for instance, the Law of Excluded Middle (LEM) p ∨ ¬p is no longer a theorem. My question: aside from the logic formal perspective, is IPL supposed to model/address some specific "kind of world" ? Thanks.
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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