Q.Prove the Boolean identity : $(A+B)(A+C) = A + BC$.

Tracing the signal through the NOT gate, the second gate, and the final NAND-shaped gate, and simplifying with De Morgan's theorem, gives $Y=\bar{A}+B+\bar{C}$.

Working

Stage 1 -- NOT gate on A: output $=\bar{A}$.

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The expression $A+AB$ simplifies to $A$ by the Boolean absorption law.

Using the distributive law, factor $A$ out of the expression: $A+AB = A(1+B)$

In Boolean algebra, the OR of any variable with logic 1 is always 1: $1+B=1$ (the identity/annulment law for OR), regardless of the value of $B$.

Substituting this back: $A(1+B) = A\cdot 1 = A$ (since $A\cdot 1 = A$, the identity law for AND).

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The input values that satisfy $\overline{(A+B)}\cdot\overline{(A\cdot B)} = 1$ are $A=0,\ B=0$.

For the product of the two complemented (NOR and NAND) terms to equal 1, BOTH terms must individually equal 1.

• $\overline{(A+B)} = 1$ requires $A+B = 0$. Since $A$ and $B$ are Boolean (0 or 1), OR gives 0 only when BOTH inputs are 0. So this term forces $A=0,\ B=0$.
• Substituting $A=0, B=0$ into the second term: $\overline{(A\cdot B)} = \overline{0\cdot 0} = \overline{0} = 1$. Both required conditions hold simultaneously, so the full product is $1\cdot 1 = 1$. ✓ …