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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:

AB'C*(BD+CDE)+AC'
Applying Distributive law:
AB'CBD + AB'CCDE + AC'
Applying A*A'=0 and C*C=C
AB'CDE+AC'
Factoring:
A*(B'CDE+C')
This is where I'm stuck. This is where I tried most of the rules and got nowhere

The books solution however is:

A*(C'+B'DE)

Then, I found this site and it also gave me a different solution. I don't know if the book has a wrong solution, or an alternate one

Boolean Algebra - Digital Electronics

Their solution was:

A * C'

Any suggestions? Thanks.

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    \$\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 '18 at 4:30
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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.

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  • \$\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\$ – CircAnalyzer Sep 20 '18 at 4:53
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    \$\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\$ – user287001 Sep 20 '18 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\$ – CircAnalyzer Sep 21 '18 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\$ – user287001 Sep 21 '18 at 8:33

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