sols1 - M114S, Solutions to HW #1 x1.1. Suppose that x A (...

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Unformatted text preview: M114S, Solutions to HW #1 x1.1. Suppose that x A ( B C ). Then either x A (in which case x belongs to both A B and A C and hence to their intersection), or else x B C (so, again, x ( A B ) ( A C ). This proves half of what we want. For the opposite direction, suppose x ( A B ) ( A C ). Then we have two cases: if x A , then x A ( B C ); and if x / A , then x must belong to both B and to C . So again, x A ( B C ). We can also combine the two arguments in symbolized form: x A ( B C ) x A or x ( B C ) x A or [ x B and x C ] [ x ( A B ) and x ( A C )] or [ x ( A B ) and ( A C )] x ( A B ) ( A B ) . The second property A ( B C ) = ( A B ) ( A C ) is proved similarly. Finally, we check that A \ ( A B ) = A \ B . If x A \ ( A B ), then x A , but x / A B , and so x cannot belong to B : therefore, x A \ B . For the opposite direction, if x A \ B , then x A and x / B , so, certainly, x / A B ; hence x A \ ( A B ). x1.2. First, let x C \ ( A B ). Then x belongs to C but not to A B ; so x belongs to neither A nor B ; thus, x belongs to C \ A and to C \ B , i.e., x ( C \ A ) ( C \ B ). Fortheoppositedirection,suppose x ( C \ A ) ( C \ B ); then x C , but also x is not in A and not in B ; so x / A B ; hence x C \ ( A B ). This verifies the first law. For the second law, we again argue both directions. Let x C \ ( A B ); then x belongs to C but not to A B ; so either x does not belong to A , or x does not belong to B , and in either case, we see that x ( C \ A ) ( C \ B )....
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This homework help was uploaded on 04/16/2008 for the course MATH 114S taught by Professor Moschovakis during the Winter '08 term at UCLA.

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sols1 - M114S, Solutions to HW #1 x1.1. Suppose that x A (...

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