# Boolean algebra simplification question

I'm trying to simplify the follow SOP expression: $\bar{A}$$\bar{B}$$\bar{C}$ + $\bar{A}$B$\bar{C}$ + $\bar{A}$BC + AB$\bar{C}$

Using a K-map (unless I've erred) it should simplify to: $\bar{A}$$\bar{C} + B\bar{C} + \bar{A}B However, I can't figure out how to get there. Here's my work: # ---------- Original expression: \bar{A}$$\bar{B}$$\bar{C} + \bar{A}B\bar{C} + \bar{A}BC + AB\bar{C} Factor \bar{A}$$\bar{C}$ out of 1st and 2nd terms: $\bar{A}$$\bar{C}(\bar{B} + B) + \bar{A}BC + AB\bar{C} Applying law of tautology to \bar{B} + B: \bar{A}$$\bar{C}$(1) + $\bar{A}$BC + AB$\bar{C}$

Applying identity to 1st term: $\bar{A}$$\bar{C} + \bar{A}BC + AB\bar{C} From that point I'm not sure how to proceed. I thought about factoring B out of the 1st and 3rd terms but that didn't clarify anything for me. I'm very new to this, having just learned it, so I reckon it's an elementary mistake that is otherwise obvious. BTW, this is not homework. I'm just studying for an exam and found this expression on the internet somewhere. Thanks! • I figured it out. I'll answer my own question as soon as the time limit is up! – CSpadawan Mar 25 '14 at 23:31 ## 1 Answer I figured it out and it was indeed an elementary mistake. I merely overlooked where I could factor. Here's the rest of the work (following where I left off in the main question): Factor \bar{C} out of 1st and 3rd terms: \bar{C}(\bar{A} + AB) + \bar{A}BC Applying identity to \bar{A} + AB: \bar{C}(\bar{A} + B) + \bar{A}BC Distributing terms: \bar{C}$$\bar{A}$ + $\bar{C}$B + $\bar{A}$BC

Factor $\bar{A}$ out of 1st and 3rd terms: $\bar{A}$($\bar{C}$ + BC) + $\bar{C}$B

Applying identity to $\bar{C}$ + BC: $\bar{A}$($\bar{C}$ + B) + $\bar{C}$B

Distributing terms: $\bar{A}$$\bar{C} + \bar{A}B + \bar{C}B Which is equivalent to what I got from the K-map: \bar{A}$$\bar{C}$ + B$\bar{C}$ + $\bar{A}$B