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?
Complement of the difference of the sets
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$\begingroup$
elementary-set-theory
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0Yes. Have you drawn a Venn diagram? – 2011-06-28
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0@Theo: Sure. As for me they illustrate, they don't prove. – 2011-06-28
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0But surely you have never doubted any result they illustrate. – 2011-06-28
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0@Gortaur: For me Venn diagram also prove - since I view them as a truth table, just organized in a different way: http://img684.imageshack.us/img684/7494/venn3tab.jpg – 2011-07-06
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0@Martin: thanks. How can you be sure for 4 sets that you didn't miss anything? – 2011-07-06
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0@Gortaur: I agree that truth tables might be easier to check for large number of sets (although you can make a mistake in it, too). To check that your diagram contains all necessary regions you just have to check that you have one region for each row of the corresponding truth table. However, my point was that the truth tables and Venn diagrams can be understood as two representations of the same thing. – 2011-07-06
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0@Martin: that's certainly right. I never used the truth tables though. Maybe I will start now. – 2011-07-06
2 Answers
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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, $$ (A - B)^c = (A \cap B^c )^c = A^c \cup (B^c )^c = A^c \cup B. $$