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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? – Berci Oct 9 '12 at 14:58
up vote 0 down vote accepted

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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Thanks man! I get it! So my solution is correct, yes? – Maximus Programus Oct 9 '12 at 14:58

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