The problem involves simplifying a logic function in Product of Sums (POS) form using algebraic manipulations and drawing its corresponding logic circuit. Let’s break it down:

Problem Analysis

The function $f(x_1, x_2, x_3)$ is given as: $f(x_1, x_2, x_3) = \prod(M_3, M_4, M_6)$ where $\prod$ represents the Product of Sums, and $M_3, M_4, M_6$ are maxterms (terms where the function output is 0).

Step 1: Write the Maxterms

The binary indices for $M_3, M_4, M_6$ correspond to truth table rows where the function equals 0. Here’s how we convert them:

• $M_3 = (x_1 + x_2' + x_3)$
• $M_4 = (x_1 + x_2 + x_3')$
• $M_6 = (x_1' + x_2 + x_3')$

So the POS expression becomes: $f(x_1, x_2, x_3) = (x_1 + x_2' + x_3)(x_1 + x_2 + x_3')(x_1' + x_2 + x_3')$

Step 2: Simplify the Expression

Now we simplify the POS expression using Boolean algebra:

1. Distribute terms to combine them: $f(x_1, x_2, x_3) = (x_1 + x_2' + x_3)(x_1 + x_2 + x_3') \cdot (x_1' + x_2 + x_3')$

   This is already in simplified POS form, so no further reduction is required for the algebraic expression.

Step 3: Draw the Logic Circuit

The POS logic circuit is constructed using:

• OR gates for the individual terms (maxterms).
• A single AND gate to combine the outputs of all the OR gates.

Steps to Draw:

1. For each maxterm:
   • $M_3 = (x_1 + x_2' + x_3)$: Connect $x_1$, $x_2'$ (NOT $x_2$), and $x_3$ to an OR gate.
   • $M_4 = (x_1 + x_2 + x_3')$: Connect $x_1$, $x_2$, and $x_3'$ (NOT $x_3$) to another OR gate.
   • $M_6 = (x_1' + x_2 + x_3')$: Connect $x_1'$ (NOT $x_1$), $x_2$, and $x_3'$ to a third OR gate.
2. Combine the outputs of the three OR gates using a single AND gate.

This forms the complete POS circuit.

Would you like a detailed diagram of the circuit, or further simplification explanations? Let me know!

Follow-Up Questions:

1. What is the difference between SOP (Sum of Products) and POS (Product of Sums) logic forms?
2. How can you convert a POS expression into a truth table?
3. What are maxterms and minterms in Boolean algebra?
4. Can this expression be simplified further using Karnaugh maps (K-maps)?
5. How does simplifying a Boolean expression reduce circuit complexity?

Tip:

To simplify complex Boolean expressions efficiently, use Karnaugh maps—they visually group minterms/maxterms to eliminate variables systematically.