1. **Problem Statement:** Convert the Boolean function $F = \Sigma(1,4,5,6,7)$ to Product of Sums (POS) form and simplify using Karnaugh map. 2. **Formula and Rules:** - Sum of Minterms (SOM) represents the function as a sum of minterms. - Product of Sums (POS) is the product of maxterms. - To convert SOM to POS, use the complement of minterms to find maxterms. - Karnaugh map helps simplify Boolean expressions by grouping 1s (for SOP) or 0s (for POS). 3. **Step 1: Identify variables and minterms** Assuming 3 variables $x, y, z$ (since max minterm is 7): - Minterms: 1 (001), 4 (100), 5 (101), 6 (110), 7 (111) 4. **Step 2: Write the function in SOP form:** $$F = m_1 + m_4 + m_5 + m_6 + m_7$$ 5. **Step 3: Find maxterms (zeros) for POS:** - Minterms not included: 0 (000), 2 (010), 3 (011) - Maxterms correspond to these zeros. 6. **Step 4: Write maxterms:** - $M_0 = x + y + z$ - $M_2 = x + y' + z$ - $M_3 = x + y' + z'$ 7. **Step 5: POS form:** $$F = (x + y + z)(x + y' + z)(x + y' + z')$$ 8. **Step 6: Karnaugh Map simplification:** | yz \ x | 0 | 1 | |-------|---|---| | 00 | 0 | 1 | | 01 | 0 | 1 | | 11 | 1 | 1 | | 10 | 0 | 1 | Group zeros (0s) at positions 0,2,3: - Group 1: cells 0 and 2 (x=0,y=0,z=0 and x=0,y=1,z=0) - Group 2: cell 3 alone (x=0,y=1,z=1) 9. **Step 7: Simplify maxterms from groups:** - Group 1 maxterm: $(x + z)$ - Group 2 maxterm: $(x + y' + z')$ 10. **Step 8: Final simplified POS:** $$F = (x + z)(x + y' + z')$$ **Answer:** The POS form of $F$ is $$(x + y + z)(x + y' + z)(x + y' + z')$$ and simplified POS is $$(x + z)(x + y' + z')$$.