Simplify the Boolean expression S = A.B.C + A.B.C, where the dot is AND and the plus is OR.

Step-by-step solution

1. 1

   Read the operators as logic, not arithmetic. In Boolean algebra $\cdot$ is AND and $+$ is OR. Neither variable is complemented here (there is no overbar on any letter), so the two product terms are identical: both are the single minterm $A\cdot B\cdot C$.

2. 2

   Identify the applicable law. For OR, the idempotent law states

   $X+X=X$

   This holds because OR asks "is at least one true?" — asking the same question twice adds nothing. Both possible values confirm it: $0+0=0$ and $1+1=1$.

3. 3

   Apply it with X = A.B.C.

   $S=A\cdot B\cdot C+A\cdot B\cdot C=A\cdot B\cdot C$

4. 4

   Contrast with ordinary algebra, where the answer would differ. In arithmetic, $x+x=2x$. In Boolean algebra there is no $2$ — the only values are $0$ and $1$, and $1+1=1$, not $2$. Writing $2ABC$ here would be meaningless.

5. 5

   Verify with a truth table. Enumerating all eight combinations of $(A,B,C)$, the expression $A\!\cdot\!B\!\cdot\!C+A\!\cdot\!B\!\cdot\!C$ equals $A\!\cdot\!B\!\cdot\!C$ in every row ✓ — output $1$ only for $A=B=C=1$, and $0$ in the other seven rows.

6. 6

   Note the practical consequence. In a circuit, the unsimplified form would use two identical three-input AND gates feeding an OR gate; the simplified form needs a single AND gate and no OR gate at all. Spotting duplicate minterms is the first pass of any gate-count reduction — and the same duplication often hides a transcription error in the original expression, worth checking before building.