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I have the following boolean expression that I want to simplify

$$B\cdot D+ \overline{A\cdot B\cdot D} + \overline{B}\cdot C\cdot \overline{D}$$

Here is what I have been able to due so far

$$B\cdot D+ \overline{A} + \overline{B} + \overline{D} + \overline{B} \cdot \overline{D}\cdot C$$

I know that the answer is suppose to be $$\overline{A} + \overline{B}\cdot B $$

How can I simplify my initial expression any further thank you very much any help.

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Factorising $$\overline{A}+B\cdot D + \overline{B}+\overline{D} \cdot (1+\overline{B}\cdot C)$$ $$=\overline{A}+B\cdot D + \overline{B}+\overline{D} \cdot 1$$ $$=\overline{A}+B\cdot D + \overline{B}+\overline{D}$$ Then by the absorption law, $$B\cdot D + \overline{B}=D + \overline{B}$$ So this simplifies to $$=\overline{A}+D + \overline{B}+\overline{D}$$ $$=\overline{A} + \overline{B}+D+\overline{D}$$ $$=\overline{A} + \overline{B}+1$$ $$=1$$

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  • $\begingroup$ I drew a Karnaugh map and all the squares were full so I agree with the answer of $1$. $\endgroup$ – poetasis Mar 22 '19 at 16:47

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