Please help me simplify this formula by using boolean algebra rules:
$F= x_1'x_2'x_3'x_4'+x_1'x_2'x_3x_4+x_1'x_2x_3'x_4'+x_1'x_2x_3x_4'+x_1x_2'x_3x_4.$
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.
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 $$x_1'x_3'x_4' \vee x_2'x_3x_4 \vee x_1'x_2x_3x_4' = (x_1'x_3'x_4'\vee x_1'x_2x_3'x_4') \vee x_2'x_3x_4 \vee x_1'x_2x_3x_4'$$ $$ = x_1'x_3'x_4' \vee x_2'x_3x_4 \vee (x_1'x_2x_3x_4'\vee x_1'x_2x_3'x_4')$$ $$ = x_1'x_3'x_4' \vee x_2'x_3x_4 \vee x_1'x_2x_4'$$