To construct the Karnaugh Map (K-map) and determine the implicants and prime implicants for the given Boolean function $F(x, y, z) = xz' + xyx + yz'$, follow these steps:

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Step 1: Simplify the Function

First, simplify the given function:

$F(x, y, z) = xz' + xyx + yz'$

1. Combine $xyx$ to $xy$ because $x \cdot x = x$: $F(x, y, z) = xz' + xy + yz'$

2. The simplified function becomes: $F(x, y, z) = xz' + xy + yz'$

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Step 2: Express Function in Minterms

To construct the K-map, rewrite $F(x, y, z)$ in terms of minterms. Identify the minterms by evaluating $xz' + xy + yz'$ for all combinations of $x, y, z$:

1. $xz'$: This is true when $x = 1$ and $z = 0$, independent of $y$.

   • Minterms: $(x, y, z) = (1, 0, 0), (1, 1, 0)$ → $m4, m6$.

2. $xy$: This is true when $x = 1$ and $y = 1$, independent of $z$.

   • Minterms: $(x, y, z) = (1, 1, 0), (1, 1, 1)$ → $m6, m7$.

3. $yz'$: This is true when $y = 1$ and $z = 0$, independent of $x$.

   • Minterms: $(x, y, z) = (0, 1, 0), (1, 1, 0)$ → $m2, m6$.

Combine all minterms: $F(x, y, z) = m2 + m4 + m6 + m7$

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Step 3: Fill the K-map

The K-map is a 3-variable grid (2x4) for $x, y, z$. Assign minterms as follows:

$yz \backslash x$	0	1
$00$	$m0$ (0)	$m4$ (1)
$01$	$m1$ (0)	$m5$ (0)
$11$	$m3$ (0)	$m7$ (1)
$10$	$m2$ (1)	$m6$ (1)

The $1$'s correspond to the minterms of $F(x, y, z)$: $m2, m4, m6, m7$.

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Step 4: Identify Implicants

Group adjacent $1$'s to form implicants:

1. Group $m4$ and $m6$: These correspond to $xz'$ (covers $100$ and $110$).
2. Group $m2$ and $m6$: These correspond to $yz'$ (covers $010$ and $110$).
3. Group $m6$ and $m7$: These correspond to $xy$ (covers $110$ and $111$).

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Step 5: Prime Implicants

Prime implicants are the largest possible groups of adjacent $1$'s that cannot be combined further:

1. Prime Implicant 1: $xz'$ (covers $m4, m6$).
2. Prime Implicant 2: $yz'$ (covers $m2, m6$).
3. Prime Implicant 3: $xy$ (covers $m6, m7$).

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Final K-map and Prime Implicants

$yz \backslash x$	0	1
$00$	$0$	$1$ ($xz'$)
$01$	$0$	$0$
$11$	$0$	$1$ ($xy$)
$10$	$1$ ($yz'$)	$1$

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Summary:

• Implicants: $xz', xy, yz'$
• Prime Implicants: $xz', yz', xy$

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Would you like me to dive deeper into essential implicants or draw the final minimized Boolean equation for you? 😊

5 Related Questions:

1. How can essential prime implicants be identified in a K-map?
2. What is the significance of redundant implicants in simplifying Boolean functions?
3. How does the size of the K-map change with more variables?
4. Can K-maps handle "don't care" conditions? If yes, how?
5. How does a K-map compare to the Quine-McCluskey method?

Tip: Always aim to group the largest number of adjacent 1's in the K-map for efficient minimization.