Let's find the minimum (simplest) Boolean expression for each function **step-by-step**. --- ## (a) \( f(A,B,C) = m_1 + m_3 + m_4 + m_6 \) ### Step 1: Write minterms in binary - \( m_1 = 001 \) (\(A'B'C\)) - \( m_3 = 011 \) (\(A'BC\)) - \( m_4 = 100 \) (\(AB'C'\)) - \( m_6 = 110 \) (\(ABC'\)) ### Step 2: Write the expression \[ f(A,B,C) = A'B'C + A'BC + AB'C' + ABC' \] ### Step 3: Simplification Group terms: - \(A'B'C + A'BC = A'C(B' + B) = A'C\) - \(AB'C' + ABC' = AC'(B' + B) = AC'\) **Final Answer:** \[ \boxed{f(A,B,C) = A'C + AC'} \] --- ## (b) \( f(d,e,f) = \sum m(1,4,5,7) \) ### Step 1: Write minterms in binary - \( m_1 = 001 \) (\(d'e'f\)) - \( m_4 = 100 \) (\(de'f'\)) - \( m_5 = 101 \) (\(de'f\)) - \( m_7 = 111 \) (\(def\)) ### Step 2: Write the expression \[ f(d,e,f) = d'e'f + de'f' + de'f + def \] ### Step 3: Simplification Group similar terms: - \(de'f' + de'f = de'(f'+f) = de'\) - \(de' + def = d(e' + ef)\) (but let's check Karnaugh map for minimization) Using K-map for three variables: | de \ f | | 1 | |--------|---|---| | 00 | | 1 (m1) | | 01 | | | | 10 | 1 (m4) | 1 (m5) | | 11 | | 1 (m7) | From the map: - \(de'f\) and \(de'f'\) combine to \(de'\) - \(def\) is on its own - \(d'e'f\) is on its own So, the minimal expression is: \[ \boxed{f(d,e,f) = de' + def + d'e'f} \] --- ## (c) \( f(r,s,t) = r't' + rs' + rs \) ### Step 1: Write the expression as is: \[ f(r,s,t) = r't' + rs' + rs \] ### Step 2: Simplification - \(rs' + rs = r(s' + s) = r\) So, \[ f(r,s,t) = r + r't' \] ### Step 3: \(r + r't' = r + t'\) Because \(r + r't' = (r + t')\), using consensus theorem. **Final Answer:** \[ \boxed{f(r,s,t) = r + t'} \] --- ## (d) \( f(a,b,c) = m_3 + m_4 + m_6 + m_7 \) ### Step 1: Write minterms in binary - \(m_3 = 011\) (\(a'bc\)) - \(m_4 = 100\) (\(ab'c'\)) - \(m_6 = 110\) (\(abc'\)) - \(m_7 = 111\) (\(abc\)) ### Step 2: Write the expression \[ f(a,b,c) = a'bc + ab'c' + abc' + abc \] ### Step 3: Simplification Group \(abc' + abc = ab\) So, \(ab'c' + ab = a(b'c' + b)\) But \(b'c' + b = (b' + b)(c' + b) = 1 \cdot (c' + b) = c' + b\), but let's check with K-map. K-map for three variables (a, b, c): | ab \ c | | 1 | |--------|---|---| | 00 | | | | 01 | | 1 (m3) | | 10 | 1 (m4) | | | 11 | 1 (m6) | 1 (m7) | From the map: - \(a'bc\) (m3) - \(ab'c'\) (m4) - \(abc'\) (m6) - \(abc\) (m7) Group \(abc' + abc = ab\) So, \[ f(a,b,c) = ab + ab'c' + a'bc \] Can't combine further. **Final Answer:** \[ \boxed{f(a,b,c) = ab + ab'c' + a'bc} \] --- ## (e) \( f(n,p,q) = \sum m(2,3,5,7) \) ### Step 1: Write minterms in binary - \( m_2 = 010 \) (\(n'pq'\)) - \( m_3 = 011 \) (\(n'pq\)) - \( m_5 = 101 \) (\(np'q\)) - \( m_7 = 111 \) (\(npq\)) ### Step 2: Write the expression \[ f(n,p,q) = n'pq' + n'pq + np'q + npq \] ### Step 3: Simplification Group \(n'pq' + n'pq = n'p\) Group \(np'q + npq = nq\) **Final Answer:** \[ \boxed{f(n,p,q) = n'p + nq} \] --- ## **Summary Table** | Part | Minimum Expression | |------|---------------------------------| | (a) | \( A'C + AC' \) | | (b) | \( de' + def + d'e'f \) | | (c) | \( r + t' \) | | (d) | \( ab + ab'c' + a'bc \) | | (e) | \( n'p + nq \) |