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Can someone show me the steps in getting from $f = (ab + c')(d' + e + f')$ to $f = abe +ab(df)' + c'e + c'(df)'$? I am trying to relearn Boolean algebra after a long hiatus.

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$$ \begin{align} f & = (ab + c')(d'+e + f') \tag{1}\\ \\ f & = (ab + c')(e + (d' + f')) \tag{2} \\ \\ f & = (ab + c')(e+(df)') \tag{3}\\ \\ f & = abe+ ab(df)' + c'e + c'(df)' \tag{4} \end{align}$$



$(3) \;\;d'+f' = (df)': \text{DeMorgan's}$

$(4)\;\; \text{Distributive Law}$

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  • $\begingroup$ Very clean and so nicely lined up! +1 $\endgroup$
    – Amzoti
    Commented May 7, 2013 at 0:40

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