My Math Forum Set theory: simplify algebraic expressions

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 August 28th, 2017, 12:34 PM #1 Member   Joined: Sep 2014 From: Sweden Posts: 94 Thanks: 0 Set theory: simplify algebraic expressions Hi! I'm new to the set theory and its algebraic expressions. So I don't really know how to apply the laws that exist. If we have the simple law: One of the commutative laws: $\displaystyle \displaystyle A \cup B = B \cup A$ Does this law apply to complements also? If we have $\displaystyle \displaystyle \mathsf{c}A$ for example? Or are laws that not include complement invalid for sets that don't are complements? My problem is this (the last expression is supposed to be A ⋃ (B - A) OBS! not the answer!): $\displaystyle \displaystyle A \cup (\mathsf{c}A \cap B) =$ $\displaystyle \displaystyle A \cup (B \cap \mathsf{c}A ) =$ $\displaystyle \displaystyle A \cup (B - A) = ?$ (I know you don't actually write B - A) Then I don't know how to go on from there. Last edited by skipjack; August 28th, 2017 at 12:47 PM.
 August 28th, 2017, 12:53 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,301 Thanks: 1971 The ⋃ and ⋂ operations are each commutative (for any pair of sets). What other rules could you use?
 August 28th, 2017, 01:53 PM #3 Global Moderator   Joined: May 2007 Posts: 6,684 Thanks: 659 B-A is equivalent to B∩cA.
 September 1st, 2017, 04:12 AM #4 Member   Joined: Sep 2014 From: Sweden Posts: 94 Thanks: 0 Could you show me an easy expression to simplify? Because I still not really get it . Not too easy expression though. Nevermind, actually clicked now Last edited by DecoratorFawn82; September 1st, 2017 at 04:33 AM.

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