Boolean Algebra – Simplification Methods

boolean-algebra

Here's the Karnaugh map:

alt text

The answer I should be getting from the Karnaugh should be:

T = R ∙ (CGM)'

I'm really not seeing how this was arrived at through any simplification methods I've learned thus far. I can see the answers that are intended are correct (I think), though.

From what I know, the best answer I can come up with to simplify (only) the Kargnaugh map is:

T=RCG'+RCM'+RC'
  to:
T=R∙(CG'+CM'+C')

Help appreciated!

Best Answer

Using sum of products you should be able to derive:

RC' + RCG' + RCGM'

Substitute out the R:

R(C' + CG' + CGM')

Use the identity A + A'B = A + B

R(C' + G' + CGM')

That same identity works as a multivariable expression

R(C' + G' + M')

Then apply DeMorgans

R(CGM)'

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