Step 1: Analyze and Rewrite the Expression

We aim to simplify $Y = ((AB)' + A' + AB)'$. First, let's simplify step-by-step:

1. $AB$: The conjunction of $A$ and $B$.
2. $(AB)'$: The negation of $AB$, equivalent to $\overline{AB}$.
3. $(AB)' + A' + AB$: The disjunction of $\overline{AB}$, $\overline{A}$, and $AB$.
4. $Y = ((AB)' + A' + AB)'$: The negation of the above expression.

Step 2: Construct the Expression for K-map

The original expression involves three variables: $A$, $B$, and $AB$. We will treat $Y$ as a function of $A$ and $B$.

Intermediate Expression:

$W = (AB)' + A' + AB$ The truth table for this can be derived as follows:

$A$	$B$	$AB$	$(AB)'$	$A'$	$W$
0	0	0	1	1	1
0	1	0	1	1	1
1	0	0	1	0	1
1	1	1	0	0	1

Finally, negate $W$ to get $Y$:

$A$	$B$	$W$	$Y = W'$
0	0	1	0
0	1	1	0
1	0	1	0
1	1	1	0

Step 3: K-map for $Y$

K-map Layout:

The Karnaugh map for $Y$ (as a function of $A$ and $B$) is filled with the truth table values:

AB & 0 & 1 \\ \hline 0 & 0 & 0 \\ 1 & 0 & 0 \\ \end{array}$$ #### Simplified Expression: The K-map for $$Y$$ contains only $$0$$'s, meaning the expression simplifies to: $$Y = 0$$ ### Final Answer: The simplified Boolean expression is: $$Y = 0$$ --- Would you like further clarifications on K-maps? ### Related Questions: 1. How does DeMorgan's theorem help simplify Boolean expressions? 2. Can you explain how to construct a truth table for a more complex Boolean function? 3. How do K-maps simplify Boolean algebra compared to algebraic methods? 4. What are the differences between SOP and POS forms in Boolean algebra? 5. How can you minimize a 4-variable Boolean expression using K-map? ### Tip: Always verify your simplified Boolean expression using a truth table to ensure accuracy.