Simplify the following boolean expression using Boolean algebra rules, K-map, and QM: AC' + CD + AB'CD + AB(C' + D) + A'D

Understand the Problem

The question asks to simplify a given Boolean expression using Boolean algebra rules, Karnaugh maps (K-map), and the Quine-McCluskey (QM) method.

Answer

$AC' + D$

Steps to Solve

Expand the expression Expand the term $AB(C' + D)$ using the distributive property: $AC' + CD + AB'CD + ABC' + ABD + A'D$
Remove redundant terms Since $AB'CD$ implies $CD$, and can be removed, the expression becomes: $AC' + CD + ABC' + ABD + A'D$
Rearrange terms Rearrangeing the terms for better visualization: $AC' + ABC' + A'D + ABD + CD$
Simplify using absorption law Simplify $AC' + ABC'$ to $AC'$: $AC'(1 + B) = AC'$

So the expression becomes: $AC' + A'D + ABD + CD$

Factor out A from the second and third term Factor out A from $ABD$ and add and subtract ${AD}$: $AC' + A'D + A(BD+D) + CD - AD$
Simplify with Boolean Algebra Simplify $A(BD + D) = AD$ $D(B+1) = D$

The expression becomes $AC' + A'D + AD + CD$

Simplify further Because $A'D + AD = D$, the equation becomes: $AC' + D + CD$
Final Simplification Simplify $D + CD$ to $D$ $D(1+C) = D$

So the final expression is: $AC' + D$

$AC' + D$

More Information

The simplified Boolean expression is $AC' + D$. This is the most reduced form of the original expression, achieved using Boolean algebra rules such as the distributive property, absorption law, and the identity $X + XY = X$.

Tips

A common mistake is not recognizing when one term implies another, such as $AB'CD$ implying $CD$, which allows for simplification. Also, errors can occur during the expansion and factoring steps if the distributive property isn't applied correctly. Another mistake is not recognizing when a term can be absorbed into another term.

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