Question about proof of associative law for sets

AI Thread Summary
The discussion revolves around proving the associative law for sets, specifically that (A U B) U C = A U (B U C). The initial approach involves assuming an element x belongs to the left-hand side and demonstrating it also belongs to the right-hand side, establishing subset relationships. A response confirms the validity of this reasoning and suggests a more formal structure for the proof, illustrating the logical equivalences involved. The clarification emphasizes the interchangeable nature of the logical "or" in the context of set unions. This exchange highlights the challenges of proving fundamental mathematical concepts that are often taken for granted.
pamparana
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Hello,

Trying to go through Tom Apostle text on Calculus. There is an exercise about proving the associative law for sets:

So, (A U B) U C = A U (B U C)

So, if we assume x to be an element in set in left hand side, than we can say x belongs at least to either A, B or C which in turn means that x is also an element in set in right hand side and then we can say that the LHS and RHS are subsets of each other...

Is this a valid proof? I am never sure with these. It is really tricky to prove such ideas that we take for granted in every day life!

Anyway, I would be really grateful for any help you can give this old man.

/Luca
 
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Hi pamparana,

What it comes down to is that "or" (http://en.wikipedia.org/wiki/Logical_disjunction" for
((p or q) or r)
and
(p or (q or r))
are the same.
 
Last edited by a moderator:
pamparana said:
Hello,

Trying to go through Tom Apostle text on Calculus. There is an exercise about proving the associative law for sets:

So, (A U B) U C = A U (B U C)

So, if we assume x to be an element in set in left hand side, than we can say x belongs at least to either A, B or C which in turn means that x is also an element in set in right hand side and then we can say that the LHS and RHS are subsets of each other...

Is this a valid proof? I am never sure with these. It is really tricky to prove such ideas that we take for granted in every day life!

Anyway, I would be really grateful for any help you can give this old man.

/Luca

You are essentially correct. (The other post is correct too, but is really a round-a-bout way to assume exactly what you want to prove). You might see the proof of your statement organized formally this way.

<br /> \begin{align*}<br /> x \in (A \cup B) \cup C &amp; \leftrightarrow x \in (A \cup B) \text{ or } x \in C \\<br /> &amp; \leftrightarrow x \in A \text{ or } x \in B \text{ or } x \in C \\<br /> &amp; \leftrightarrow x \in A \text{ or } x \in (B \cup C) \\<br /> &amp; \leftrightarrow x \in A \cup (B \cup C)<br /> \end{align*}<br />

I've use \leftrightarrow to represent the phrase "if and only if" (I couldn't get the usual double arrow to work, sorry).
Hope this helps.
 
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...

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