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I was reading the distributive law of sets (I keep coming back to basic maths when needed, forget it after some time, then come back again. Like I'm in loop):

$A\cup(B \cap C)=(A\cup B)\cap(A\cup C)$

The proof (which I'm assuming everyone knows) has a transaction between lines which baffled me , which are:

$x \in A$ or ($x\in B$ and $x\in C$)

($x\in A$ or $x\in B$) and ($x\in A$ or $x \in C$)

in the second line, did they just applied distributive law? (in the proof of distributive law itself Oo) or Did they simple assumed "and" = "+" etc like following:

$2 X (A + B) \equiv (2XA) + (2XB)$

Another question will be :

($x\in A$ or $x\in B$) or $x\in C\implies x\in A$ or ($x\in B$ or $x\in C$)

I can just open the brackets?

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What's relation between your first and second examples? This example seems irrelevant here: $2 X (A + B) == (2XA) + (2XB)$ – Gigili Sep 16 '12 at 17:54
@Gigili I edited the question, sorry for ma mistake – Mr.Anubis Sep 16 '12 at 18:02
up vote 5 down vote accepted

It does seem quite naughty to do this sort of thing; but yes, you are allowed. But only if you know what you're doing.

However, it does not assume the result. $\cap$ and $\cup$ are set operations, whereas 'and' and 'or' are logical operations. More precisely, in logic, we can write $p \wedge q$ to mean '$p$ and $q$' and $p \vee q$ to mean '$p$ or $q$'. By use of truth-tables, we can prove things like $p \vee (q \wedge r) \leftrightarrow (p \vee q) \wedge (p \vee r)$, and these proofs have nothing specifically to do with set theory. But then these logical rules transfer across by setting $p$ to be the assertion $x \in A$, $q$ to be the assertion $x \in B$, and so on.

$+$ and $\times$ are arithmetic operators and thus yet another thing altogether from $\cap$,$\cup$ and $\wedge$,$\vee$.

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I still don't understand how could make transaction like that Oo also you overlooked my second question I guess :) – Mr.Anubis Sep 16 '12 at 18:01
Sorry I thought it was clear $-$ you can just open the brackets; 'and' distributes over 'or' and 'or' distributes over 'and'. But don't see it as a purely algebraic rule like this $-$ think it over with some examples so that you know it makes sense. – Clive Newstead Sep 16 '12 at 18:22
your answer is somewhat right but still I need to know the rules, solid statements, might be you could redirect me somewhere where I can read about them in more details. I'd like you to emphasize on line of my question : did they just applied distributive law? (in the proof of distributive law itself Oo) or Did they simple assumed "and" = "+" etc like following: – Mr.Anubis Sep 16 '12 at 21:04
@Mr.Anubis you keep saying "the distributive law". The distributive law on what? They did apply a distributive law on logical operations in a proof of a distributive law on set operations, but this isn't circular because they aren't the same thing. – Robert Mastragostino Sep 16 '12 at 21:14
@Robert: You put very concisely what I tried to say in my answer :) Is it clearer now, @MrAnubis? – Clive Newstead Sep 16 '12 at 22:03

I think your problem understanding what's going on here stems from the fact that you don't quite understand what's the mathematical logic/set theory operations behind "everyday" words:

$\,x\in A\cup\left(B\cap C\right)\,$ means "either $\,x\,$ belongs to $\,A\,$ or else $\,x\,$ belongs to $\,B\cap C\,$ (why the "either"? Because that's the meaning of union of sets). Also remember that in mathematics "or" is not exclusive: it may also mean "both".

Now, "$\,x\,$ belongs to $\,B\cap C\,$" means "$\,x\,$ belongs to $\,B\,$ and also $\,x\,$ belongs to $\,C\,$" , so putting all these together we get that $\,x\in A\cup\left(B\cap C\right)\,$ means "either $\,x\,$ belongs to $\,A\,$ or $\,x\,$ belongs to both $\,B\,\,and\,\,C\,$" , which is the same, both mathematically and colloquially, as "$\,x\,$ belongs to $\,A\,$ and $\,B\,$ or $\,x\,$ belongs to $\,A\,$ and $\,C\,$" , and this last is just the wordly meaning of the right hand side in that equality.

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