Boolean Simplification Problem

Question

I have to simplify the figure

I came up with the equation $$(ac+\bar b)\cdot (a+b+\bar c) \cdot (b+ac).$$

Help me out

Attempted Answer:

\begin{align*} (ac+\bar b)\cdot (a+b+c)\cdot (b+ac) &=(ac+abc+ac+a\bar b+\bar b c)\cdot(b+ac)\\ &=abc+abc+abc+ac+abc+ac+a\bar b c+a\bar b c \\ &=abc+a\bar b c+ac\\ &=ac(b+\bar b)+ac\\ &=ac+ac=ac \end{align*}

Answer 1

OK, here goes:

We need to simplify

$(ac + \bar b)(a + b + \bar c)(b + ac). \tag{1}$

I'll do this by a method which is a little different than the OP's, but will yield the same answer $ac$; this will corroborate k31453's work. First, I observe that

$(ac + \bar b)(a + b + \bar c)(b + ac) = (ac + \bar b)(ac + b)(a + b + \bar c), \tag{2}$

and we have

$(ac + \bar b)(ac + b) = (ac)(ac) + acb + ac \bar b + \bar b b = ac + acb + ac \bar b, \tag{3}$

since $(ac)(ac) = ac$ and $\bar b b = 0$. Next, I use

$acb + ac \bar b = ac(b + \bar b) = ac1 = ac \tag{4}$

to obtain

$(ac + \bar b)(ac + b) = ac + ac = ac. \tag{5}$

Substituting this back into (2) yields

$ac(a + b + \bar c) = ac + abc = ac(1 + b) = ac1 = ac, \tag{6}$

confirming k31453's calculation.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!