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I have to solve this simple boolean function :

$$f_1 * f_2 = (x_1 + x_2) * (!x_1 + x_3)$$

The solution is : $x_1*x_3 + !x_1*x_2$

Can anyone make a step by step solution because after getting :

$$x_1*x_3 + x_2*(!x_1) + x_2*x_3$$

I don't know how to proceed (I got stuck).

Thanks

GoodDeeds
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Alfonso
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2 Answers2

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If $x_1$ then you get $x_3+x_2*x_3 = x_3(1+x_2) = x_3$; if $!x_1$ then you get $x_2+x_2*x_3 = x_2(1+x_3) = x_2$. In either case, $x_2*x_3$ is redundant, so it can be discarded from the solution.

abiessu
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As abiessu says the $x_2*x_3$ term is redundant. But the actual theorem is called consensus. $X Y + \overline X Z + Y Z = X Y + \overline X Z$

$$x_1*x_3 + x_2*\overline x_1 + x_2*x_3$$

Another way of looking at it. Expand term 3 and have it absorbed in the two other terms.

$$x_1*x_3 + x_2*\overline x_1 + x_2*x_3*(\overline x_1 +x_1)$$ $$x_1*x_3 + x_2*\overline x_1 + \overline x_1 *x_2*x_3* +\ x_1*x_2*x_3$$ Look for common terms. $$x_1*x_3 * (1 + x_2) + \overline x_1* x_2 * (1 + x_3)$$ $$x_1*x_3 + \overline x_1* x_2$$

StainlessSteelRat
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