You now have a Venn diagram for A ∪ B ∪ C
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A or B or C = A + B + C - A and B - A and C - B and C - 2 (A and B and C) I'm not sure by the way;
Draw your Venn Diagram as three overlapping circles. Each circle is a set. The union of the sets is what's contained within all 3 circles, making sure not to count the overlapping portion twice. An easier problem is when you have 2 sets, lets say A and B. In a Venn Diagram that looks like 2 overlapping circles. A union B = A + B - (A intersect B) A intersect B is the region that both circles have in common. You subtract that because it has already been included when you added circle A, so you don't want to add that Again with circle B, thus you subtract after adding B. With three sets, A, B, C A union B union C = A + B - (A intersect B) + C - (A intersect C) - (B intersect C) + (A intersect B intersect C) You have to add the middle region (A intersect B intersect C) back because when you subtract A intersect C and B intersect C you are actually subtracting the very middle region Twice, and that's not accurate. This would be easier to explain if we could actually draw circles.
not (b or c) = (not b) and (not c)
No- this is not true in general. Counterexample: Let a = {1,2}, b = {1} and c ={2}. a union c = [1,2} and b union c = {1,2} but a does not equal b. The statement be made true by putting additional restrictions on the sets.
complement of c