2) Consider a6 -variable Boolean function f=f_(1)(x_(1),x_(2),x_(3),x_(4))cdot f_(2)(x_(4),x_(5),x_(3)) where f_(1)=prod (1,2,3,5,7,12,14)-X_(1) is the most significant bit, and f_(2)=prod (3,4,5,6,7)-X_(4) is the most significant bit. a) Obtain a minimal product-of-sum (POS)expression for f. b) Implement f using only two-input NAND (NAND-2) gates; use minimal number o gates. Use only variables as inputs (not their negated forms).

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2) Consider a6 -variable Boolean function f=f_(1)(x_(1),x_(2),x_(3),x_(4))cdot f_(2)(x_(4),x_(5),x_(3)) where f_(1)=prod (1,2,3,5,7,12,14)-X_(1) is the most significant bit, and f_(2)=prod (3,4,5,6,7)-X_(4) is the most significant bit. a) Obtain a minimal product-of-sum (POS)expression for f. b) Implement f using only two-input NAND (NAND-2) gates; use minimal number o gates. Use only variables as inputs (not their negated forms).

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a) To obtain a minimal product-of-sum (POS) expression for f, we need to find the prime implicants for each function $f_{1}$ and $f_{2}$. For $f_{1}$, the given expression is $f_{1}=\prod (1,2,3,5,7,12,14)-X_{1}$. This means that $f_{1}$ is equal to 1 for all combinations of $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, x_{7}, x_{8}$ except when $x_{1}$ is 1. The prime implicants for $f_{1}$ are: $A_{1} = \overline{x_{1}}$ $A_{2} = x_{2}x_{3}x_{4}x_{5}x_{6}x_{7}x_{8}$ $A_{3} = x_{1}x_{2}x_{3}x_{4}x_{5}x_{6}x_{7}x_{8}$ $A_{4} = x_{1}x_{2}x_{3}x_{4}x_{5}x_{6}x_{7}x_{8}$ $A_{5} = x_{1}x_{2}x_{3}x_{4}x_{5}x_{6}x_{7}x_{8}$ $A_{6} = x_{1}x_{2}x_{3}x_{4}x_{5}x_{6}x_{7}x_{8}$ $A_{7} = x_{1}x_{2}x_{3}x_{4}x_{5}x_{6}x_{7}x_{8}$ $A_{8} = x_{1}x_{2}x_{3}x_{4}x_{5}x_{6}x_{7}x_{8}$ For $f_{2}$, the given expression is $f_{2}=\prod (3,4,5,6,7)-X_{4}$. This means that $f_{2}$ is equal to 1 for all combinations of $x_{3}, x_{4}, x_{5}, x_{6}, x_{7}$ except when $x_{4}$ is 1. The prime implicants for $f_{2}$ are: $B_{1} = \overline{x_{4}}$ $B_{2} = x_{3}x_{4}x_{5}x_{6}x_{7}$ $B_{3} = x_{3}x_{4}x_{5}x_{6}x_{7}$ $B_{4} = x_{3}x_{4}x_{5}x_{6}x_{7}$ $B_{5} = x_{3}x_{4}x_{5}x_{6}x_{7}$ $B_{6} = x_{3}x_{4}x_{5}x_{6}x_{7}$ $B_{7} = x_{3}x_{4}x_{5}x_{6}x_{7}$ $B_{8} = x_{3}x_{4}x_{5}x_{6}x_{7}$ Now, we can write the POS expression for $f$ as: $f = \overline{x_{1}} \cdot (x_{3}x_{4}x_{5}x_{6}x_{7}) + \overline{x_{4}} \cdot (x_{2}x_{3}x_{4}x_{5}x_{6}x_{7}x_{8})$ b) To implement $f$ using only two-input NAND (NAND-2) gates, we can use the following steps: 1. Create a NAND-2 gate with inputs $x_{1}$ and $x_{3}$. The output of this gate will be $\overline{x_{1}}$. 2. Create a NAND-2 gate with inputs $x_{4}$ and $x_{5}$. The output of this gate will be $\overline{x_{4}}$. 3. Create a NAND-2 gate with inputs $x_{2}$ and $x_{3}$. The output of this gate will be $\overline{x_{2}x_{3}}$. 4. Create a NAND-2 gate with inputs $x_{4}$ and $x_{5}$. The output of this gate will be $\overline{x_{4}x_{5}}$. 5. Create a NAND-2 gate with inputs $x_{6}$ and $x_{7

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