I'm currently stuck on the expression listed below. I applied almost all the Boolean algebra rules trying to figure it out, but no luck

Here is the expression $$\boxed{\mathtt{AB'C\cdot(BD+CDE)+AC'}}$$

  • Applying Distributive law
    AB'CBD + AB'CCDE + AC'

  • Applying A⋅A'=0 and C⋅C=C

  • Factoring

This is where I'm stuck. This is where I tried most of the rules and got nowhere

The Textbook Solution however is: $$\boxed{\mathtt{A⋅(C'+B'DE)}}$$

Then, I found this site and it also gave me a different solution. Their solution was $$\boxed{\mathtt{A \cdot C'}}$$

I don't know if the book has a wrong solution, or an alternate one.

Any suggestions? Thanks.

  • 1
    \$\begingroup\$ Your book's answer is right. The web site you linked doesn't like "E". (Just look at the table they generate -- it is missing a column for E.) So just replace "E" with "F" on that web site and see what answer it gives you then. Note that it generates a proper table, too, if you use F instead of E. \$\endgroup\$
    – jonk
    Sep 20, 2018 at 4:30

1 Answer 1


This is a common case. I have had a possiblity to see, that quite many do not intuitively see the following identity: X'+XY = X'+Y. Learn it.

AC' cannot be equivalent with your original formula. AC'=1 only if A=1 and C=0. Your original formula=1 in a case where C=1.

  • \$\begingroup\$ The book we are using didn't present that Identity rule. Instead it presented the following 12 rules: 1. A+0=A, 2.A+1=1, 3.A*0=0, 4.A*1=A, 5. A+A=A, 6. A+A'=1, 7. AA=A, 8. AA'=0, 9.A''=A, 10. A+AB = A, 11. A+A'B = A+B, 12. (A+B)(A+C) = A+BC. I guess the Identity you referenced is similar to rule 11, but not quite the same....??? Anyhow, I was able to manipulate it by using Demorgans Law, rule 11, simplification, and demorgans law again to get to the answer... \$\endgroup\$ Sep 20, 2018 at 4:53
  • 1
    \$\begingroup\$ @DeeTee rule 11 is the same as I wrote, only negate X' and you get your rule 11. A variable can be negated if it's done at both sides and every instance is negated. \$\endgroup\$
    – user136077
    Sep 20, 2018 at 6:43
  • \$\begingroup\$ So do I negate all the terms in the parens then apply rule 11, or do I have to negate the whole expression, then apply rule 11? \$\endgroup\$ Sep 21, 2018 at 3:35
  • \$\begingroup\$ @DeeTee You do not negate anything in A(B'CDE+C'). You modify rule 11 to an equivalent form where negate signs fit - I have done it visibly, normally one does it in the fly. - and apply it. X=C Y=B'DE. \$\endgroup\$
    – user136077
    Sep 21, 2018 at 8:33

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