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What is $(A\setminus B)^c$? It seems to me that it is equal to $B\cup A^c$, isn't it? Please help me to verify if there is a mistake?

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Yes. Have you drawn a Venn diagram? – t.b. Jun 28 '11 at 12:36
@Theo: Sure. As for me they illustrate, they don't prove. – Ilya Jun 28 '11 at 12:38
But surely you have never doubted any result they illustrate. – André Nicolas Jun 28 '11 at 13:09
@Gortaur: For me Venn diagram also prove - since I view them as a truth table, just organized in a different way: – Martin Sleziak Jul 6 '11 at 12:18
@Martin: thanks. How can you be sure for 4 sets that you didn't miss anything? – Ilya Jul 6 '11 at 13:14
up vote 4 down vote accepted

Using the property $A \backslash B = A \cap B^c$ and the De-Morgan's laws you get the following chain of equalities:

$$ (A \backslash B)^c = (A \cap B^c)^c = A^c \cup (B^c)^c = A^c \cup B $$

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Yes, $$ (A - B)^c = (A \cap B^c )^c = A^c \cup (B^c )^c = A^c \cup B. $$

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