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Can this Boolean expression:

$$A*\overline{A*B}$$

be expanded to give:

$$A*\overline{A} * A*\overline{B}$$

Although that appears to reduce to zero?

I know $A(\overline{A+B})$ can be expanded to give: $A*A + A*\overline{B}$

So can it work with an AND? How else do you simplify the first expression?

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  • $\begingroup$ To put it over multiple characters, you can use \overline{} instead of \bar{}: $w+\overline{x+y}+z$. $\endgroup$ Jan 5, 2012 at 3:05
  • $\begingroup$ A more general comment: If you work with small boolean expressions (<4 variables) a truth-table for both expressions can always determine equivalence. $\endgroup$
    – chazisop
    Jan 5, 2012 at 16:20

1 Answer 1

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No. By De Morgan's laws, $$(A * B)' = A' + B'.$$ So, $A*(A*B)'$ can be expanded to give $$A*(A'+B') = A*A' + A*B' = \mathsf{F} + A*B' = A*B'.$$

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  • $\begingroup$ you're kidding! How did I not see that??, I apologise for wasting your time :) $\endgroup$
    – Jonathan.
    Jan 4, 2012 at 23:39

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