- #1
magnifik
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I'm trying to prove
abd + a'bc + ab'c + bcd + acd = abd + a'bc + ab'c
i originally thought
abd + c(a'b + ab' + bd + ad) =
abd + c(a'b + ab' + d(a+b)) =
abd + c(a'b + ab'), where a+b = 0
but i realized if a+b = 0, ab = 0 as well, which would cancel out the abd, and that is clearly incorrect. any suggestions?? thanks
abd + a'bc + ab'c + bcd + acd = abd + a'bc + ab'c
i originally thought
abd + c(a'b + ab' + bd + ad) =
abd + c(a'b + ab' + d(a+b)) =
abd + c(a'b + ab'), where a+b = 0
but i realized if a+b = 0, ab = 0 as well, which would cancel out the abd, and that is clearly incorrect. any suggestions?? thanks