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Simplify the following Boolean expression. Show your work and list which axiom or theorem you used in each step. Your final equation should be in minimized sum-of- products (SOP) form.

M = (A + B'C)(A'D + AB + AB’C + AC')B

How can this be done?

I have: Distribute = (AB+B'CB)(A'D + AB + AB'C + AC')

Combining = (AB + C)(A'D + AB + AB'C + AC')

What could be next? I can't split the four sets in the second parenthesis as they all have different stuff. Is there any further to go?

What does it mean to put something in Minimized SOP form?

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  • $\begingroup$ Your "combining" expression is actually wrong. It should be $$AB(A'D+AB+AB'C+AC'),$$ giving a different final expression. $\endgroup$ – amrsa Feb 28 '18 at 12:35
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Normally you have to set up a Karnaugh map to work out this kind of problem but in this case $$\begin{align}(A+\bar BC)(\bar AD+AB+A\bar BC+A\bar C)B&=(A+\bar BC)B(\bar AD+AB+A\bar BC+A\bar C)\\ &=(AB+B\bar BC)(\bar AD+AB+A\bar BC+A\bar C)\\ &=(AB+0C)(\bar AD+AB+A\bar BC+A\bar C)\\ &=(AB+0)(\bar AD+AB+A\bar BC+A\bar C)\\ &=AB(\bar AD+AB+A\bar BC+A\bar C)\\ &=A\bar ABD+AABB+AAB\bar BC+AAB\bar C\\ &=0BD+AB+A0C+AB\bar C\\ &=0+AB+0+AB\bar C\\ &=AB+AB\bar C\\ &=AB1+AB\bar C\\ &=AB(1+\bar C)\\ &=AB1\\ &=AB\end{align}$$ Where we have used the commutative and, associative, and distributive laws. Also that facts that $0A=0$, $1A=1$, $0+A=A$, $1+A=1$, $AA=A$. and $A\bar A=0$. You can check by verifying that if $A=B=1$ the expression evaluates to $1$ while if $A=0$ or $B=0$ the expression evaluates to $0$.

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Starting boolean expression is: (A+B'C)(A'D+AB+AB'C+AC')B

Now using Distributive law (A+B'C)(A'BD+ABB+AB'BC+ABC')

Now using idempotent law and complimentarity law i.e AA'=0 and AA=A respectively we get

(A+B'C)(A'BD+AB+ABC')

Using distributive law again: AA'BD+AAB+AABC'+A'BB'CD+ABB'C+ABB'CC'

Again after application of idempotent law and complimentarity law we have: AB+ABC' Now taking common AB we get: AB(1+C')

As 1+anything=1(in boolean algebra) we get: AB.1

Again as 1.A=A, therefore final answer is AB

Which is required SOP form. SOP means sum of product form.Like in the expression AB+ABC' which has products(i.e. AB , ABC') which are summed together.

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By distributivity, you expressions becomes $$ABA'D+AB+ABB'C+ABC'+A'CD+ABC+AB'C+ACC',$$ which, applying complements is $$AB+ABC'+A'CD+ABC+AB'C,$$ and by absorption, $$AB+A'CD+ABC+AB'C$$ Now \begin{align} AB+A'CD+ABC+AB'C &= AB + (ABC + AB'C) + A'CD\\ &= AB + AC(B+B') + A'CD\\ &= AB + AC + A'CD\\ &= AB + AC + ACD + A'CD\\ &= AB + AC + (A+A')CD\\ &= AB + AC + CD. \end{align} Above, we use $AC=AC+ACD$, which is an instance of absorption.


This is actually wrong because I picked on an intermediary expression derived by the asker in the question and that expression was wrong, so I hould have started with $$ABA'D+AB+ABB'C+ABC'.$$

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