0

Variables are $E, B, S, V_1, V_2, V_3$

Here is the Boolean expression that I need to simplify

$$EB'SV_1'V_2'V_3 + EB'SV_1'V_2V_3 + EB'SV_1V_2'V_3 + EB'SV_1V_2V_3' + EB'SV_1V_2V_3$$

And this is how I simplified,

$$\begin{align} &EB'SV_1'V_2'V_3 + EB'SV_1'V_2V_3 + EB'SV_1V_2'V_3 + EB'SV_1V_2V_3' + EB'SV_1V_2V_3 \\ &EB'S(V_1'V_2'V_3 + V_1'V_2V_3 + V_1V_2'V_3 + V_1V_2V_3' + V_1V_2V_3) \\ &EB'S[V_2'V_3(V_1 + V_1') + V_2V_3(V_1 + V_1') + V_1V_2V_3'] \\ &V_1 + V_1' = 1 \end{align}$$

So,

$$\begin{align} &EB'S[V_2'V_3 + V_2V_3 + V_1V_2V_3'] \\ &EB'S[V_3(V_2 + V_2') + V_1V_2V_3'] \\ &V_2 + V_2' = 1 \end{align}$$

So,

$$EB'S(V_3 + V_1V_2V_3')$$

But when I solve this equation through a simulator, it gives me the simplified expression as

$$EB'S(V_3 + V_1V_2)$$

Can anyone tell me where did I do wrong?

Daniel P
  • 2,710
Nureka Rodrigo
  • 3
  • 1
  • Please use mathjax! $V_1$ gives $V_1;$ better: avoid the subscripts and use $P,Q,R$ instead. – ryang Jul 07 '22 at 12:11

1 Answers1

1

You have done nothing wrong, you just didn't see an opportunity to simplify a bit. The two expressions within parentheses are actually equivalent:

$$\begin{align} V_3 + V_1V_2 &= V_3 + V_1V_2(1) \\ &= V_3 + V_1V_2(V_3 + V_3') \\ &= (1)V_3 + V_1V_2V_3 + V_1V_2V_3' \\ &= (1 + V_1V_2)V_3 + V_1V_2V_3' \\ &\stackrel{(*)}{=} (1)V_3 + V_1V_2V_3' \quad\\ &=V_3 + V_1V_2V_3' \end{align}$$

$(*)$ note that this step is sort of unique to Boolean algebra and not too obvious

Daniel P
  • 2,710
PMar
  • 26
  • 1