# How can i simplyfy this boolean equation?

$F= x_1'x_2'x_3'x_4'+x_1'x_2'x_3x_4+x_1'x_2x_3'x_4'+x_1'x_2x_3x_4'+x_1x_2'x_3x_4.$

I know that the answer should be: $(x_1'x_3'x_4')+(x_1'x_2x_4')+(x_2'x_3x_4)$ from Karnaugh map. I also put it in logic friday and it simplified to this. I have so far tried to factor out the common characters, but it didnt seem to lead anywhere near the final answer.

$=x_1'x_2'(x_3'x_4'+x_3x_4)+x_1'x_2(x_3'x_4'+x_3x_4')+x_1x_2'x_3x_4$

Also noticed another way: $x_3'x_4'(x_1'x_2+x_1'x_2')+x_3x_4(x_1'x_2'+x_1x_2')+x_1'x_2x_3x_4'$

My guess is that i would need to get my equation to a form where i can use the law of absorption somehow to get rid of something, but how is the question.

• Was a mistake in third operative. was:$x_1'x_2x_3'x_4$, but actually must be:$x_1'x_2x_3'x_4'$ – rauno45 Jan 1 '16 at 16:23
• Is $x' = \not x$? That is, $0' = 1$ and $1' = 0$? – Paul Sinclair Jan 1 '16 at 17:47
• No, x' is not x. – rauno45 Jan 1 '16 at 17:51

Your supposed answer is only correct if you are interpreting (counter to standard convention) that "+" means disjunction. Conventionally, it represents exclusive-or (combining with conjuction to form the two-element field $\Bbb F_2$). Disjunction is respresented by the symbol $\vee$. The variance occurs when $x_2 = 1$, while the others are $0$. In this case $F = 1$, while $$(x_1'x_3'x_4')\text{ xor }(x_1'x_2x_4')\text{ xor }(x_2'x_3x_4) = 1 \text{ xor } 1\text{ xor } 0 = 0$$ and $$(x_1'x_3'x_4')\vee(x_1'x_2x_4')\vee(x_2'x_3x_4) = 1 \vee 1\vee 0 = 1$$
No matter which is meant by "+", it is true that $x' + x = 1$. Also, $$x_1'x_2'x_3'x_4'+x_1'x_2x_3'x_4' = x_1'(x_2 + x_2')x_3'x_4' = x_1'x_3'x_4'$$ and similarly $$x_1'x_2'x_3x_4+x_1x_2'x_3x_4 = x_2'x_3x_4$$
So the original expression reduces to $$F = x_1'x_3'x_4' + x_2'x_3x_4 + x_1'x_2x_3x_4'$$
The remaining reduction only applies to $\vee$: Note that $x_1'x_3'x_4' = x_1'x_3'x_4' \vee x_1'x_2x_3'x_4'$, so $$x_1'x_3'x_4' \vee x_2'x_3x_4 \vee x_1'x_2x_3x_4' = (x_1'x_3'x_4'\vee x_1'x_2x_3'x_4') \vee x_2'x_3x_4 \vee x_1'x_2x_3x_4'$$ $$= x_1'x_3'x_4' \vee x_2'x_3x_4 \vee (x_1'x_2x_3x_4'\vee x_1'x_2x_3'x_4')$$ $$= x_1'x_3'x_4' \vee x_2'x_3x_4 \vee x_1'x_2x_4'$$