Computer Science 2214A/B Study Guide - Final Guide: University Of Western Ontario, Distributive Property, Disjoint Sets

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The rst one is short, but requires some intuition and practice, which can be gained by drawing pictures, like the one below. The second solution is longer and less elegant; but it is also more detailed. B a c b, which implies b = a = c. The rst hypothesis, namely a b = c implies that both a and b are contained in c, that is. Now looking at the third hypothesis, namely (a c) b = a, we see that by applying distributivity of over we obtain: (a b) (c b) = a. Now using distributivity of over , we further deduce: Since c b is contained in c, we have c (c b) = c and we derive this equality: It remains to be proved that a = b holds. Using c = a within the the second hypothesis, namely (a c) b = c, we obtain: that is, (a a) b = a,