Talk:Union Distributes over Intersection

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Should instead show different two cases:

1. For EVERY X, x in S U X is in X so in intersection, so x in S U intersection (not just ONE X)
2. at least one X does not have x in X so has x in S so x in S U intersection --Arthur 21:21, 2 July 2011 (CDT)
Amended as suggested. Good call. --prime mover 03:58, 3 July 2011 (CDT)