I know the answer is AC + ~AB, but how?

I have tried:

B(~A+C) + AC and stop.

Also, I have tried:

AC + BC + ~AB and have stopped.

There seems nowhere to go in either case.

  • 4
    \$\begingroup\$ Show what you've tried and be explicit with where you are confused. Help us to help you. \$\endgroup\$
    – Bort
    Mar 1, 2019 at 18:37

1 Answer 1


I will not give you the full solution, but the required non-trivial sidestep. You have got this far:

C(A + B) + A'B = AC + BC + A'B =

Now here is the sidestep. We know that (A+A') is 1, so we can do:

 = AC + (A + A')BC + A'B 

From here you will need to expand it and use the "OR absorption law" twice.

  • \$\begingroup\$ @AlanKazemian (Be sure and hit the check mark so you can close the question) \$\endgroup\$
    – user103380
    Mar 1, 2019 at 19:11
  • \$\begingroup\$ @Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem. \$\endgroup\$
    – Pzy
    Mar 1, 2019 at 21:00

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