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How can we simplify $$F=MNO+Q'P'N'+PRM+Q'OMP'+MR$$ using the theorems of boolean algebra, not Karnaugh or anything else?

Well, I can obviously simplify $MR(P+1)=MR$, so the expression becomes $$MNO+Q'P'N'+MR+Q'OMP'$$ But from here, I tried to use De Morgan or to calculate the negative form of $F$, but none of this helps.

Iulia
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  • What have you tried? What theorems/rules do you think you might need to apply? Why don't you include what you've managed to do so far. – amWhy Mar 09 '13 at 16:30
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    Consider incorporating your above comment into your question. It will help keep others from telling you things you already know. – user642796 Mar 09 '13 at 16:45
  • Thanks, I did this. Sorry, I'm new to this site... – Iulia Mar 09 '13 at 16:49

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$$\begin{align*} F&=MNO+Q′P′N′+PRM+Q′OMP′+MR\\ &=MNO+Q′P′N′+Q′OMP′+MR(P+1)\\ &=MNO+Q′P′N′+Q′OMP′+MR\\ &=MNO+MR+Q'P'N'+Q'OMP'\\ &=MNO+MR+Q'P'N'+(Q'P'N')'Q'OMP'\\ &=MNO+MR+Q'P'N'+(Q+P+N)Q'OMP'\\ &=MNO+MR+Q'P'N'+NQ'OMP'\\ &=MNO(1+Q'P')+MR+Q'P'N'\\ &=MNO+MR+Q'P'N' \end{align*}$$

Brian M. Scott
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xavierm02
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