q = f(B) = the probability that a randomly selected zygote in a population will have allele B

1 = p + q

1

^{2}= (p + q)

^{2}= p

^{2}+ 2pq + q

^{2}

f(A)

^{2}= f(AA) = the probability that a randomly selected organism in a population with have the genotype AA

f(B)

^{2}= f(BB) = the probability that a randomly selected organism in a population with have the genotype AA

f(AB) = the probability that a randomly selected organism in a population with have the genotype AA

1 = f(AA) + f(AB) + f(BB)

p

^{2}+ 2pq + q

_{2}= f(AA) + f(AB) + f(BB)

f(A)

^{2}+ 2pq + f(B)

^{2}= f(A)

^{2}+ f(AB) + f(B)

^{2}

2pq = f(AB)

The only things I did to derive the above equations is square both sides of an equation, substitute equivalent variables, and subtract identical amounts from both sides of an equation. Mathematics says that doing any of these things to an equation CANNOT change the validity* of an equation.

I was just told that iff a population is at Hardy-Weinberg Equilibrium, then all of the above is true. That means that there are populations where the above is not true. In other words, values exist that are counterexamples to equations above, which in turn means that these mathematical truths are false. Because a mathematical truth cannot be false, all populations must be in Hardy-Weinberg Equilibrium. The whole point of the Hardy-Weinberg Equilibrium is that an evolving population will not be in an Hardy-Weinberg Equilibrium.

What am I missing?

*What I mean by this is that iff the original equation is true, then the derived equation is true.