Solving for abd + a'bc + ab'c = abd + a'bc + ab'c + bcd + acd

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The discussion centers on proving the equation abd + a'bc + ab'c + bcd + acd = abd + a'bc + ab'c. The original poster attempts to simplify the left side but realizes that assuming a + b = 0 leads to incorrect cancellations. Another participant points out that the equation appears flawed due to extra terms on the left side that are not present on the right. This discrepancy suggests that the original equation may not hold true as stated. Further clarification or alternative approaches are needed to resolve the issue.
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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
 
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magnifik said:
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

The equation doesn't look right. The terms on the right are repeated on the left, but there are extra terms on the left as well. How could that work? That's probably why you are getting some forced zeros...
 
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