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Here’s the Karnaugh map:

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

- 'Algebraic' way to prove the boolean identity $a + \overline{a}*b = a + b$
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```
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!

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- proving logical equivalence $(P \leftrightarrow Q) \equiv (P \wedge Q) \vee (\neg P \wedge \neg Q)$

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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