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$\begingroup$
A = A U '('A U B) 

I know how to prove A U '('A U B) but how do I prove the other side? Is this correct? Is it ok just to add 'B?

A /* identity laws */ A U Ø /*domination laws */ A U (Ø ∩ 'B) /* distributive laws */ (A U Ø) ∩ ( A U 'B) /*identity laws */ A ∩ ( A U 'B) /* distributive laws */ (A ∩ A) U (A ∩ 'B) /*idempotent laws */ A U (A ∩ 'B) /*de morgan's laws */ A U '('A U B) 
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    The same applies backwards, no?2012-10-09

1 Answers 1

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You have that $A = A \cup X$ for some set $X$. You'd hope that $X \subseteq A$ so RHS is not bigger than LHS.

$(A' \cup B)' = A \cap B' \subseteq A$ so we're good.

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    Thank$s$ man! I get it! So my $s$olution i$s$ correct, ye$s$?2012-10-09