Question

Prove the De Morgan’s Theorems using a truth table.

Hint:

Remember: {Break the bar, change the gate}. NOT(AND) → OR with inverted inputs, NOT(OR) → AND with inverted inputs.

Answer 1

Concept: De Morgan’s Theorems are fundamental Boolean algebra laws:

• Complement of AND = OR of complements
• Complement of OR = AND of complements

We prove both using truth tables. Theorem 1: \((A \cdot B)' = A' + B'\)}
Step 1: Construct the truth table. \[ \begin{array}{|c|c|c|c|c|} \hline A & B & A \cdot B & (A \cdot B)' & A' + B' \\ \hline 0 & 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 \\ 1 & 0 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 & 0 \\ \hline \end{array} \] Observation: Columns \( (A \cdot B)' \) and \( A' + B' \) are identical. Hence proved. Theorem 2: \((A + B)' = A' \cdot B'\)}
Step 2: Construct the truth table. \[ \begin{array}{|c|c|c|c|c|} \hline A & B & A + B & (A + B)' & A' \cdot B' \\ \hline 0 & 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 \\ \hline \end{array} \] Observation: Columns \( (A + B)' \) and \( A' \cdot B' \) are identical. Hence proved. Conclusion:
Both De Morgan’s Theorems are verified using truth tables: \[ (A \cdot B)' = A' + B', \quad (A + B)' = A' \cdot B' \] Thus, the complement of AND becomes OR of complements, and vice versa.