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If I have the boolean equation:

H = M'CD' + MC + MC' + CRD

I think I can combine so that it's

H = M'CD' + M(C + C') + CRD

Does C + C' go to simplify to zero? So, I'm left with

H = M'CD' + CRD

user1766888
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    No, $C+C'$ simplifies to $1$, so $MC + MC'$ simplifies to $M$. Recall that the proposition "(I am sitting on a chair) or (I am not sitting on a chair)" is a tautology. – Adriano Aug 09 '13 at 03:25
  • Thought so. That's what wikipedia was telling me too. So the final equation would be H = M'CD' + M + CRD? Anymore simplification to be done here? – user1766888 Aug 09 '13 at 03:32

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Karnaugh maps are your best friend. Otherwise, observe that:

$$ \begin{align*} H &= M'CD' + M + CRD \\ &= M'CD' + M(1) + CRD \\ &= M'CD' + M(CD'+1) + CRD \\ &= M'CD' + MCD' + M + CRD \\ &= (M' + M)CD' + M + CRD \\ &= (1)CD' + M + CRD \\ &= CD' + M + CRD \\ &= CD' + CRD + M \\ &= C(1)D' + CRD + M\\ &= C(1 + R)D' + CRD + M\\ &= CD' + CRD' + CRD + M\\ &= CD' + CR(D' + D) + M\\ &= CD' + CR(1) + M\\ &= CD' + CR + M\\ \end{align*} $$

Adriano
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  • Actually, I used K maps to get the first boolean equation in my original post but I'm just not sure it's correct. http://math.stackexchange.com/questions/463314/deriving-truth-table-from-english-description – user1766888 Aug 09 '13 at 04:20
  • Could you also please help me on this problem: I have everything :( http://math.stackexchange.com/questions/463404/modify-boolean-equation-to-get-3-input-nor-equation-using-boolean-algebra-rules – user1766888 Aug 09 '13 at 09:55