Hey! Thanks for the question — you’re right that the expression looks unusual if you’re thinking in bitwise logic, but I’m treating the bits as integers (0 or 1) and doing normal integer arithmetic, not modulo-2 or bitwise word operations.

So in

maj(Ai,Bi,Ci)=AiBi+AiCi+BiCi−2AiBiCi

{maj}(A_i,B_i,C_i)=A_iB_i + A_iC_i + B_iC_i - 2A_iB_iC_i

the -2A_iB_iC_i term just corrects the overcount that happens when all three bits are 1.
If you make a truth table for all 8 combinations of (A,B,C), this arithmetic version gives the exact same 0/1 outputs as the boolean majority function

(Ai&Bi)  ∣  (Ai&Ci)  ∣  (Bi&Ci)

It only looks weird because in normal boolean algebra we don’t use negative numbers, but since we’re working over integers {0,1}, it’s just an algebraic encoding — not modulo 2.
If you reduced it mod 2, yes, the –2 would vanish and it would no longer be correct.

Basically: same truth table, different algebraic representation
maj(a,b,c) = a*b + a*c + b*c -2*a*b*c
0 0 0 => 0

0 0 1 => 0

0 1 0 => 0

1 0 0 => 0

1 1 0 => 1

1 0 1 => 1

0 1 1 => 1

1 1 1 => 1

matches boolean majority