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I'm working through an example that looks like a fairly simple Karnaugh map and simplification, but I feel stupid that I can't seem to understand the correct answer.

This is the map:
K-Map

My groupings:
K-Map grouping

This is the equation:
Eq

But I don't know how it simplifies to the correct answer here:
Eq

I understand why the NOT A goes outside of the bracket, but I don't see how the C OR NOT C just becomes a C inside the bracket?

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  • $\begingroup$ Welcome to Computer Science! Note that you can use LaTeX here to typeset mathematics in a more readable way. See here for a short introduction. $\endgroup$ – Raphael Nov 20 '14 at 10:26
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    $\begingroup$ Create a full table for both formulae. Are they equivalent? $\endgroup$ – Raphael Nov 20 '14 at 10:27
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enter image description here

Check for solution. Grouping should be done using 2 quads

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    $\begingroup$ Note that you can use Markdown and LaTeX here to typeset mathematics, rather than using an image. See here for a short introduction to Latex. $\endgroup$ – D.W. Dec 30 '15 at 1:02
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First, when you are trying to group, you must use the biggest possible group. A group's size has to be a power of 2 (1, 2, 4, 8...).

In this case you need to make 2 groups:

cells 4, 5, 12, 13: $A'D$
cells 8, 9, 12, 13: $A'C$

Represented as SOP: $A'D+A'C$

Using the Distributive Law:

$A'D+A'C=A'(C+D)$


About the way that you tried:

You grouped:

cells 4, 5: $A'C'D$
cells 8, 9, 12, 13: $A'C$

Assuming that you didn't notice the bigger grouping option, the expression that you have can still be simplified using Boolean Laws:

$A'C'D+A'C=A'(C'D+C)$ Distributive Law
According to one of the laws: $A+A'B=A+B$
So: $A'(C'D+C)=A'(C+D)$

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