Which one of the following gives the simplified sum of products expression for the Boolean function F = m₀ + m₂ + m₃ + m₅, where m₀, m₂, m₃ and m₅ are minterms corresponding to the inputs A, B and C and A as the MSB and C as the LSB?

**$\text{Boolean}$ Function Simplification with K-map**

To find the simplified sum of products (SOP) expression for the given Boolean function $F=m_0+m_2+m_3+m_5$, we will use a 3-variable Karnaugh Map (K-map). The inputs are A, B, and C, with A as the Most Significant Bit (MSB) and C as the Least Significant Bit (LSB).

**$\text{Minterm}$ to Binary Conversion**

First, let's convert the given minterms into their binary representations and corresponding literal forms for inputs A, B, and C:

- $m_0$ corresponds to binary $000$, which is represented as $\bar{A}\bar{B}\bar{C}$.
- $m_2$ corresponds to binary $010$, which is represented as $\bar{A}B\bar{C}$.
- $m_3$ corresponds to binary $011$, which is represented as $\bar{A}BC$.
- $m_5$ corresponds to binary $101$, which is represented as $A\bar{B}C$.

So, the Boolean function in its canonical SOP form is:

$F=\bar{A}\bar{B}\bar{C}+\bar{A}B\bar{C}+\bar{A}BC+A\bar{B}C$

**$\text{K-map}$ Construction and Population**

Now, we will construct a 3-variable K-map and place a '1' in the cells corresponding to these minterms. A 3-variable K-map has $2^3=8$ cells.**$\text{Grouping}$ '1's for Simplification

Now we group adjacent '1's in the K-map to find the largest possible groups (prime implicants), aiming to cover all the '1's with the minimum number of groups.

- Group 1:** Group $m_0$ and $m_2$. - $m_0=\bar{A}\bar{B}\bar{C}$ (000) - $m_2=\bar{A}B\bar{C}$ (010) - These two minterms share $\bar{A}$ and $\bar{C}$. The variable B changes from $\bar{B}$ to $B$, so it gets eliminated. - The simplified term for this group is $\bar{A}\bar{C}$. This is an essential prime implicant because $m_0$ is uniquely covered by this group.
- Group 2: Group $m_5$. - $m_5=A\bar{B}C$ (101) - This minterm cannot be grouped with any other adjacent '1' to form a larger group of 2, 4, or 8. - The simplified term for this group is $A\bar{B}C$. This is also an essential prime implicant because $m_5$ is uniquely covered by this group.
- Group 3: Cover the remaining uncovered minterm $m_3$. - $m_3=\bar{A}BC$ (011) - Minterm $m_3$ can be grouped with $m_2$ (which is already covered by Group 1, but can be reused). - $m_2=\bar{A}B\bar{C}$ (010) - These two minterms share $\bar{A}$ and $B$. The variable C changes from $\bar{C}$ to $C$, so it gets eliminated. - The simplified term for this group is $\bar{A}B$.

**$\text{Simplified}$ Sum of Products Expression**

By combining all the prime implicants that cover all the minterms, we get the simplified sum of products expression:

$F=\bar{A}\bar{C}+\bar{A}B+A\bar{B}C$

This expression ensures that all original minterms ($m_0,m_2,m_3,m_5$) are covered and is in its minimal SOP form.

**$\text{Comparing}$ with Options**

Let's compare our derived simplified expression with the given options:

- Option 1: $\bar{\mathrm{A}}\mathrm{B}+\bar{\mathrm{A}}\bar{B}\bar{\mathrm{C}}+\mathrm{A}\bar{\mathrm{B}}\mathrm{C}$ (Incorrect)
- Option 2: $\bar{\mathrm{A}}\bar{\mathrm{C}}+\bar{\mathrm{A}}\mathrm{B}+\mathrm{A}\bar{\mathrm{B}}\mathrm{C}$ (Correct)
- Option 3: $\bar{\mathrm{A}}\bar{\mathrm{C}}+\mathrm{A}\bar{\mathrm{B}}+\mathrm{A}\bar{\mathrm{B}}\mathrm{C}$ (Incorrect)
- Option 4: $\bar{\mathrm{A}}\mathrm{BC}+\bar{\mathrm{A}}\bar{\mathrm{C}}+\mathrm{A}\bar{\mathrm{B}}\mathrm{C}$ (Incorrect)

The simplified expression matches Option 2.