M mous99 Oct 2012 32 0 malaysia Oct 22, 2012 #1 For any sets A and B, use algebraic laws of sets to simplify (a) [ A - (A-B)] - B (b) A - [ B - (A-B)]

For any sets A and B, use algebraic laws of sets to simplify (a) [ A - (A-B)] - B (b) A - [ B - (A-B)]

B Bingk Aug 2009 170 36 Oct 22, 2012 #2 Use \(\displaystyle X \setminus Y = X \cap Y^c\) to convert the set minus into intersection/union of sets Reactions: 1 person

Use \(\displaystyle X \setminus Y = X \cap Y^c\) to convert the set minus into intersection/union of sets

M mous99 Oct 2012 32 0 malaysia Oct 22, 2012 #3 Bingk said: Use \(\displaystyle X \setminus Y = X \cap Y^c\) to convert the set minus into intersection/union of sets Click to expand... How to put the set minus into the set?(Cool)

Bingk said: Use \(\displaystyle X \setminus Y = X \cap Y^c\) to convert the set minus into intersection/union of sets Click to expand... How to put the set minus into the set?(Cool)

B Bingk Aug 2009 170 36 Oct 22, 2012 #4 I don't understand your question ... Example: \(\displaystyle X \setminus ( Y\setminus Z) = X \ ( Y \cap Z^c) = X \cap (Y \cap Z^c)^c = X \cap (Y^c \cup Z) = \\ (X \cap Y^c) \cup (X \cap Z)\) The last line is from distribution (you can check Algebra of sets - Wikipedia, the free encyclopedia for more "shortcuts")

I don't understand your question ... Example: \(\displaystyle X \setminus ( Y\setminus Z) = X \ ( Y \cap Z^c) = X \cap (Y \cap Z^c)^c = X \cap (Y^c \cup Z) = \\ (X \cap Y^c) \cup (X \cap Z)\) The last line is from distribution (you can check Algebra of sets - Wikipedia, the free encyclopedia for more "shortcuts")