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I have a boolean function. Unfortunately I am not able to simplify it to the minimal term.

What I have done so far:

$f = x_4 x_2 + \overline{(x_3 + x_2) \cdot 1 x_1} + x_2x_0$

$f =x_4 x_2 + \overline {(x_3 + x_2 )x_1} + x_2x_0$

$f = x_4x_2 + \overline {x_3} \overline {x_2}+ \overline{x_1} + x_2x_0$

$f = x_2(x_4+x_0) + \overline {x_3} \overline {x_2} + \overline {x_1}$

$f = \overline { \overline {x_2(x_4 + x_0) + \overline {x_3} \overline {x_2}+ \overline{x_1}}}$

$f =\overline {( \overline {x_2} + (\overline {x_4} \overline {x_0})) (x_3 + x_2)x_1} $

$f = \overline {(\overline {x_2} + (\overline {x_4} \overline {x_0})) x_3x_! + x_2x_!}$

The correct solution should be: $f = \overline {x_1} + \overline {x_3x_2} + x_2x_0 + x_4x_2$

Have I done a mistake or do I just don't see the next step?

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    $\begingroup$ Didn't you just show it on the third step? $\endgroup$ – projectilemotion Mar 4 '17 at 12:14
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Your deMorgans conversion from 2nd to 3rd step is incorrect.

$f = x_4 x_2 + \overline {(x_3 + x_2 )x_1} + x_2x_0$

$f = x_4 x_2 + \overline {(x_3 + x_2 )} + \overline {x_1} + x_2x_0$

$f = x_4 x_2 + \overline {x_3}\ \overline {x_2} + \overline {x_1} + x_2x_0$

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