# Boolean Algebra, Simplification: Sum of products from truth table

Based on a previous question (and this ought to be the last of this sequence I promise), I just need to know one more thing. According to this truth table:

I'm told the answer I should be getting by using the sum of products method on the truth table should be:

T = R ∙ (RCGM)'

Unlike before with the Karnaugh map, I can't see how this was arrived at. All I can get is:

T = RC'G'M'+RC'G'M+RC'GM'+RC'GM+RCG'M'+RCG'M+RCGM'
T = R(C'G'M'+C'G'M+C'GM'+C'GM+CG'M'+CG'M+CGM')
T = R(C'(G'M'+G'M+GM'+GM)+C(G'M'+G'M+GM'))
T = R(C'(G'(M'+M)+G(M'+M))+C(G'(M'+M)+GM'))
T = R(C'(G'+G)+C(G'+GM'))
T = R(C'+C(G'+M'))
T = R(C'+G'+M')
T = R(CGM)'

This is the right answer I believe, but it's not the same as what I should supposedly have found.

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The two results are the same.

R(RCGM)'=RR'+RC'+RG'+RM'=RC'+RG'+RM'=R(CGM)'

On the other hand, this means the second R is kind of redundant, so I don't know why you'd want to write the solution this way rather than your way.

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Rather I think the suggestion of $R(RCGM)'$ is a product of sums which -can- be read off directly:

• $R$ for -all- the rows where R is negative
• $(RCGM)' = R'+C'+G'+M'$ for the single row where all are true
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