Q.State Boolean laws. Elucidate how they are used to simplify Boolean expressions with suitable example.

Step 1 (The laws). Complement law: $A\cdot\overline{A}=0$, $A+\overline{A}=1$. OR laws: $A+0=A$, $A+1=1$, $A+A=A$. AND laws: $A\cdot0=0$, $A\cdot1=A$, $A\cdot A=A$. Commutative: $A+B=B+A$, $A\cdot B=B\cdot A$. Associative: $A+(B+C)=(A+B)+C$, $A\cdot(B\cdot C)=(A\cdot B)\cdot C$. Distributive: $A(B+C)=AB+AC$, and the Boolean-specific $A+BC=(A+B)(A+C)$.

Step 2 (Why they matter). A raw Boolean expression written straight from a circuit's wiring often has redundant terms; applying these laws algebraically reduces the expression to its simplest equivalent form, and since each term in the final expression corresponds to a physical gate, a shorter expression means FEWER gates needed to build the same logic function -- lower cost, less complexity, less power.

Step 3 (Worked example). Simplify $AC+ABC$. Factor out the common term AC: $AC(1+B)$. By OR law 2 ($X+1=1$, applied with $X=B$), $1+B=1$, so the expression becomes $AC\cdot1$. By AND law 2 ($X\cdot1=X$), $AC\cdot1=AC$. Therefore $AC+ABC=AC$. …

State the law families, then apply them step by step to simp …

• Forgetting which specific law justifies each simplification step (e.g. citing 'AND law' generically instead of the correct numbered law). …