Let's solve the Boolean function step by step: ### Step 1: Write the Minterms Given: \[ f(A, B, C) = \sum m(0, 2, 3, 5, 6, 7) \] Let's write each minterm in terms of \(A\), \(B\), and \(C\): | Minterm Number | A | B | C | Expression | |----------------|---|---|---|------------------| | 0 | 0 | 0 | 0 | \(A'B'C'\) | | 2 | 0 | 1 | 0 | \(A'BC'\) | | 3 | 0 | 1 | 1 | \(A'BC\) | | 5 | 1 | 0 | 1 | \(AB'C\) | | 6 | 1 | 1 | 0 | \(ABC'\) | | 7 | 1 | 1 | 1 | \(ABC\) | --- ### Step 2: Write the Canonical SOP Expression Combine all the minterms: \[ f(A,B,C) = A'B'C' + A'BC' + A'BC + AB'C + ABC' + ABC \] --- ### Step 3: Group and Factor Let's group and factor terms: #### Group \(A'B'C'\) and \(A'BC'\): \[ A'B'C' + A'BC' = A'C'(B' + B) = A'C' \] #### Group \(A'BC\): \[ A'BC \] #### Group \(AB'C\): \[ AB'C \] #### Group \(ABC'\) and \(ABC\): \[ ABC' + ABC = AB(C' + C) = AB \] Now combine all simplified terms: \[ f(A,B,C) = A'C' + A'BC + AB'C + AB \] --- ### Step 4: Check for Further Simplification Let’s look for further simplification. #### Combine \(A'BC\) and \(AB'C\): \[ A'BC + AB'C = (A'B + AB')C = (A \oplus B)C \] where \(\oplus\) denotes XOR. So, the expression becomes: \[ f(A,B,C) = A'C' + (A \oplus B)C + AB \] But let's check if we can further combine with \(AB\): \[ (A \oplus B)C + AB = (A'B + AB')C + AB \] \[ = A'BC + AB'C + AB \] But \(A'BC + AB'C + AB\) cannot be further simplified easily with \(A'C'\). So, the minimized expression is: --- ## **Final Minimized Expression** \[ \boxed{f(A,B,C) = A'C' + A'BC + AB'C + AB} \] Alternatively, you can leave it as: \[ \boxed{f(A,B,C) = A'C' + (A \oplus B)C + AB} \] Both are valid minimal forms using algebraic manipulation.

Let's carefully minimize the Boolean function \(f(A, B, C) = \sum m(0, 2, 3, 5, 6, 7)\) step-by-step with detailed calculations. --- ### **Step 1: Write the minterms in Boolean form** Given minterms: | Minterm Number | Binary (A B C) | Boolean Expression | |-----------------|----------------|------------------------------| | 0 | 000 | \(A' B' C'\) | | 2 | 010 | \(A' B C'\) | | 3 | 011 | \(A' B C\) | | 5 | 101 | \(A B' C\) | | 6 | 110 | \(A B C'\) | | 7 | 111 | \(A B C\) | --- ### **Step 2: Write the sum of minterms (Sum of Products form)** \[ f(A, B, C) = A' B' C' + A' B C' + A' B C + A B' C + A B C' + A B C \] --- ### **Step 3: Group minterms to simplify** Let's analyze the terms to find common factors. **Group 1:** \(A' B' C'\) **Group 2:** \(A' B C'\) and \(A' B C\) **Group 3:** \(A B' C\) **Group 4:** \(A B C'\) and \(A B C\) --- ### **Step 4: Simplify each group** #### **Group 2:** \(A' B C' + A' B C\) Factor out \(A' B\): \[ A' B C' + A' B C = A' B (C' + C) = A' B \times 1 = A' B \] *since \(C' + C = 1\).* #### **Group 4:** \(A B C' + A B C\) Factor out \(A B\): \[ A B C' + A B C = A B (C' + C) = A B \times 1 = A B \] --- ### **Step 5: Rewrite the function with simplified terms** Now, the function becomes: \[ f = A' B' C' + A' B + A B' C + A B \] --- ### **Step 6: Look for further simplification** Our current expression: \[ f = A' B' C' + A' B + A B' C + A B \] Let’s analyze terms: - \(A' B'\ C'\) - \(A' B\) - \(A B'\ C\) - \(A B\) --- ### **Step 7: Group terms to find common factors** **Group A:** \(A' B\) - Already simplified. **Group B:** \(A' B'\ C'\) - Isolated; no common factors with others. **Group C:** \(A B' C\) - Also isolated. **Group D:** \(A B\) - Covers all cases where \(A=1, B=1\), regardless of \(C\). But note that this includes \(A=1, B=1, C=0\) and \(A=1, B=1, C=1\), which are minterms 7 and 6 respectively. --- ### **Step 8: Construct Karnaugh Map (K-map) for confirmation** To ensure minimality, let's fill the 3-variable K-map: | **AB\C** | 00 | 01 | 11 | 10 | |----------|-------|-------|-------|-------| | **0** (A'B') | 1 (m0) | 0 | 1 (m2) | 0 | | **1** (AB) | 0 | 1 (m5) | 1 (m7) | 1 (m6) | - Minterm 0: A'B'C' → cell (A'B', C' ) → row 0, col 00 - Minterm 2: A' B C' → row 0, col 10 - Minterm 3: A' B C → row 0, col 11 (not in the minterms; note minterm 3 is 011) - Minterm 5: A B' C → row 1, col 01 - Minterm 6: A B C' → row 1, col 10 - Minterm 7: A B C → row 1, col 11 Now, fill in the minterms: | **AB\C** | 00 | 01 | 11 | 10 | |----------|-------|-------|-------|-------| | **0** (A'B') | 1 | 0 | 1 | 1 | | **1** (AB) | 0 | 1 | 1 | 1 | --- ### **Step 9: Find the largest possible groups** - **Group 1:** Cover minterms 0, 2, 6, 7 (cells at positions (A'B', C'), (A'B, C'), (AB, C'), (AB, C)) But note that minterms 6 and 7 are adjacent vertically, and minterms 0 and 2 are adjacent horizontally. - **Group 2:** Cover minterms 6 and 7 (row 1, columns 10 and 11) - **Group 3:** Cover minterms 0 and 2 (row 0, columns 00 and 10) - **Group 4:** Cover minterms 5 and 7 (row 1, columns 01 and 11) --- ### **Step 10: Write the simplified expressions from groups** From the groups: - **Group 1:** minterms 0, 2, 6, 7 - This group covers all minterms where \(A=1\) or \(A=0\), so not a simple common factor. - **Group 2:** minterms 6 and 7: - These are both with \(A=1\), \(B=1\), regardless of \(C\). - **Expression:** \(A B\) - **Group 3:** minterms 0 and 2: - Both with \(A=0\), \(B=0\), \(C=0\) or \(C=1\) (since minterm 2 has \(A' B\) and \(C=0\)) - But minterm 2 is \(A' B C'\) - So, **expression:** \(A' B'\ C'\) - **Group 4:** minterms 5 and 7: - Both with \(A=1\), \(B=0\) or \(B=1\), but minterm 5 has \(A=1, B=0, C=1\), minterm 7 has \(A=1, B=1, C=1\). - They share \(A=1, C=1\), regardless of \(B\). - **Expression:** \(A C\) --- ### **Step 11: Final simplified expression** Combining all: \[ f = A B + A' B' C' + A C \] This is a minimal sum of products expression. --- ### **### **Final Answer:** \[ \boxed{ \boxed{ f(A,B,C) = A B + A' B' C' + A C } } \] This is the minimized Boolean function obtained via algebraic simplification and confirmed through Karnaugh map grouping. --- **Summary:** - The function simplifies to three product terms. - It covers all original minterms with minimal expressions. Let me know if you'd like further explanation!