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I think I'm regularly screwing up some very simple logic and inclusion concepts in set theory revolving around and/or operators and want to figure out on which points my views are wrong.

  1. $P = \emptyset \land Q = \emptyset$. I've seen a proof showing that this means $P\cup Q = \emptyset$. While my English interpretation is that if they are independently equal to $\emptyset$, it is clear that their union must also $ = \emptyset$. Similarly to how $0 + 0 = 0$. However, mathematically I've been taught to associate $\land$ with $\cap$. Does this mean the statement $P = \emptyset \land Q = \emptyset$ is different from $P \land Q = \emptyset \implies P\cap Q = \emptyset?$

  2. $P = \emptyset \lor Q = \emptyset$. My English interpretation of this is that at least one of $P, Q$ is equal to the emptyset, but it is possible that both are (it still satisfies logical or). We can't bet on both being equal to $\emptyset$, because we're only sure 1 is. The intersection between $(ANYTHING)\cap \emptyset = \emptyset$ so this is the most general guarantee we can make. However I'm tempted to swap $\lor$ out with $\cup$ because I'm taught they are essentially equivalent. Since in this situation it seems they are not I'm lead to believe $P = \emptyset \lor Q = \emptyset$ is different from $P \lor Q = \emptyset \implies P\cup Q = \emptyset$.

I think my confusion with a lot of this comes from the fact that $and$ as well as $or$ don't necessarily act as operators in English text. For example the statement:

  • 1.) $A$ and $C$ are disjoint and 2.) $B$ and $C$ are disjoint - in English this directly translates to #1 above, however when writing it out I come up with this $\forall x, x \notin (A\cap C)\cap(B\cap C)$. Am I taking the 2nd "and" too literally, and treating it as an operator? It seems that sometimes "and" means $\cap$ but sometimes not and I'm a bit confused. I think I can make the justification that with this case, "and" does not imply any operation on the two disjoint sets because the statement never says anything about them being disjoint or not. It just talks about each set $P = A\cap C, Q = B\cap C$ as being independently disjoint.

Below on paper exist the breaking down of several logical (or illogical lol) inclusion statements. I have gotten two answers for most of them and can't figure out where my logic is wrong. Any guidance/help in clearing this stuff up ill be very very appreciated! I think I'm greatly over thinking this topic because it seems fairly straight forward and trivial.

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  • $\begingroup$ The condition $P = \emptyset$ and $Q = \emptyset$ clearly implies $P \cup Q = \emptyset$, and also $P \cap Q = \emptyset$. $\endgroup$ Sep 5 '16 at 14:43
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    $\begingroup$ $P \land Q = \emptyset$ is wrongly written, unless it is a (bad) abbreviation for : $P = \emptyset \land Q = \emptyset$. $\endgroup$ Sep 5 '16 at 14:44
  • $\begingroup$ The "analogy" between $\land$ and $\cap$ is correct, at "set level"; the def is : $x \in P \cap Q$ iff $x \in P$ and $x \in Q$. $\endgroup$ Sep 5 '16 at 14:51
  • $\begingroup$ But if we have e.g. $x \in P$ and $P \ne \emptyset$, this does not mean that we may use $\cap$. $\endgroup$ Sep 5 '16 at 14:52
  • $\begingroup$ @MauroALLEGRANZA ok I think I understand what you mean by at "set level". Can you take a look at the proofs on paper, I'm concerned with my jump from lines 3 -> 5. Given what you said, I think only the one veering to the right is valid, is this true? Because if x is not on P or x is not on Q, that means it still may be on one of them so we can't say anything for sure x being on the union. And if the one veering right is true, how come the left one seems to be intuitively wrong, but correct on paper? $\endgroup$ Sep 5 '16 at 15:43
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Before answering your main question, let me point out that (in your scanned argument) the jump from 23 to 24 is false: "$\forall x(x\not\in A\vee x\not\in B)$" does not imply "$\forall x(x\not\in A\cup B)$." For example, set $A=\{1\}, B=\{2\}$ - the first statement is true, the second is false.


There is indeed an analogy between "$\wedge$" and "$\cap$" (and similarly between "$\vee$" and "$\cup$"); however, it's more complicated than you think.

It has to do with possibilities. Think about the statements $$A=\mbox{"It is raining,"}\quad B=\mbox{"I am sick"}.$$ Now ask: what will be true tomorrow?

Well, tomorrow hasn't happened yet; so (barring determinism) we need to talk about possible futures. There is a set $[A]$ of possible futures in which $A$ is true, and a set $[B]$ of possible futures in which $B$ is true.

Now, what are the possible futures where $A\wedge B$ is true? It's exactly $[A]\cap [B]$!

This analogy takes a lot of work to make precise; one standard approach is via modal logic. What you're doing is seeing the conjunction of two statements about sets, and trying to turn that into a statement about the intersection of those sets; that doesn't work.

Here's an example that may be clearer: consider the statements "$5\in P$" and "$Q$ is infinite." It should be clear that the conjunction of these two sentences has nothing to do with $P\cap Q$. I think this is easier to see because the two statements don't even have the same form, so there's less temptation to conflate "$\cap$" and "$\wedge$" incorrectly.

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  • $\begingroup$ Regarding your preface to your answer, this means that my jump from lines 3 => 5 is also false correct? Which is why negating it is the only way to actually get the truth? $\endgroup$ Sep 5 '16 at 16:26
  • $\begingroup$ @DomFarolino The left-hand-side jump 3->5 is false, yes. (The right-hand-side jump is true.) And you can check this by setting $P=\{1\}, Q=\{2\}$. $\endgroup$ Sep 5 '16 at 16:28
  • $\begingroup$ Ok that makes sense. It seems like the negation strategy is a nice way to expose the definition of union/intersection so that we can finally swap a logical and/or with a set operator. Sorry I keep bugging you but I also assume 12->14 on the left is illegal, since it is not the definition of either. Instead we negate it to form the definition of union and then move on. Yeah I think I've been taking the conjunction of two statements and trying to force a set operation on them. However it only works out it if you can show the statement is equal to the definition of some operation. $\endgroup$ Sep 5 '16 at 17:36
  • $\begingroup$ @DomFarolino Actually 12->14 is correct, although it might be clearer to do it in more than one step. Think about what it means first, and convince yourself that it's true; then try to do it using the rules. $\endgroup$ Sep 5 '16 at 17:43
  • $\begingroup$ Logically I know its true, I just found it super weird that two different ways to do the same thing (left and right) yielded two results, both true, where one (on the right) is more general, and one (on the left) is more specific. $\endgroup$ Sep 5 '16 at 17:46

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