Let $X$ be a 3-variable Boolean function that produces output as ‘1’ when at least two of the input variables are ‘1’. Which of the following statement(s) is/are CORRECT, where $a, b, c, d, e$ are Boolean variables?

The function $X(p, q, r)$ is the 3-variable majority function:

Key properties of the majority function include:

• Idempotence / Repeated variables: $X(p, p, q) = p \cdot p + p \cdot q + q \cdot p = p + pq = p$.

• Permutation symmetry: $X(p, q, r)$ is invariant under any permutation of its arguments.

Now let us evaluate each statement:

• Option (A): Consider the assignment $a = 0, b = 0, c = 1, d = 1, e = 1$:

  $$\text{LHS} = X(0, 0, X(1, 1, 1)) = X(0, 0, 1) = 0$$

  $$\text{RHS} = X(X(0, 0, 1), 1, 1) = X(0, 1, 1) = 1$$

  Since $\text{LHS} \neq \text{RHS}$, statement (A) is incorrect.

• Option (B):Let $M = X(a, b, c) = ab + bc + ca$. Then:

  $$X(a, b, M) = ab + bM + aM = ab + (a + b)M$$

  Substituting $M = ab + bc + ca$:

  $$ab + (a + b)(ab + bc + ca) = ab + abc + abc = ab + bc + ca = X(a, b, c)$$

  Thus, statement (B) is correct.

• Option (C): Note that $X(a, b, a) = a$ and $X(c, d, c) = c$. The RHS is therefore $a \cdot c$. If we choose $a = 0, b = 1, c = 1, d = 1$:

  $$\text{LHS} = X(0, 1, X(0, 1, 1)) = X(0, 1, 1) = 1$$

  $$\text{RHS} = X(0, 1, 0) \text{ AND } X(1, 1, 1) = 0 \cdot 1 = 0$$

  Since $\text{LHS} \neq \text{RHS}$, statement (C) is incorrect.

• Option (D): Since $X(a, c, c) = c$, the expression on the RHS is:

  $$\text{RHS} = X(a, X(a, b, c), c)$$

  By symmetry, $X(a, X(a, b, c), c) = X(a, c, X(a, c, b))$. From the property verified in (B), this simplifies to $X(a, c, b) = X(a, b, c)$. Thus, statement (D) is correct.

Therefore, the correct options are B and D.