Let \(𝑋\) be a 3-variable Boolean function that produces output as ‘1’ when…

Summary of correct identities for X, the 3-variable majority function (output 1 when at least two inputs are 1):

• The equality X(a, b, X(a, b, c)) = X(a, b, c) is correct. Reason: consider cases on a and b. If a and b are both 1 the value is 1; if both 0 the value is 0; if exactly one of a and b is 1 then X(a,b,c) = c and substituting yields the same value. Thus the inner repetition does not change the majority.

• The equality X(a, b, c) = X(a, X(a, b, c), X(a, c, c)) is correct. Reason: X(a,c,c) = c because two copies of c determine the majority, so the right-hand side becomes X(a, t, c) with t = X(a,b,c). One checks that X(a,t,c) = t by a short case analysis: if a = c then t = a and both sides equal a; if a ≠ c, then t = majority(a,b,c) is the same value returned by majority(a,t,c). This follows from the idempotent/absorbing behavior of the majority operator.

• The equality X(a, b, X(c, d, e)) = X(X(a, b, c), d, e) is not correct. Counterexample: a = 0, b = 0, c = d = e = 1. Then X(c,d,e) = 1, so left side X(0,0,1) = 0, while X(a,b,c) = 0 gives right side X(0,1,1) = 1. The two sides differ.

• The equality X(a, b, X(a, c, d)) = (X(a, b, a) AND X(c, d, c)) is not correct. Simplify the right-hand side: X(a,b,a) = a and X(c,d,c) = c, so RHS = a AND c. Counterexample: a = 1, b = 0, c = 0, d = 1 gives left side X(1,0,X(1,0,1)) = X(1,0,1) = 1, while RHS = 1 AND 0 = 0.

Final answer: the correct statements are the ones asserting X(a, b, X(a, b, c)) = X(a, b, c) and X(a, b, c) = X(a, X(a, b, c), X(a, c, c)).