I'm trying to prove that $\mathrm{HALF}\text-\mathrm{FALSE}$ is NP-hard, where $\mathrm{HALF}\text-\mathrm{FALSE}$ is the following problem:
given a boolean formula $\phi(x_1,\dots,x_n)$, is there a satisfying assignament in which exactly $n/2$ variables have value false?
Now, i got the problem down if it was like this:
given a boolean formula $\phi(x_1,\dots,x_n)$, is there a satisfying assignament in which at least $n/2$ variables have value false?
the reduction for this modified problem from $\mathrm{SAT}$ to (let's call it $\mathrm{HALF}\text-\mathrm{FALSE}'$) i think is the following: $$f(\phi(x_1,\dots,x_n))=\phi(x_1,\dots,x_n)\bigwedge_{i=1}^{n}y_i'$$
it's correct because $\phi \in SAT \Leftrightarrow f(\phi) \in \mathrm{HALF}\text-\mathrm{FALSE}'$ and $f$ is computable in polynomial time.
Back to the original problem, my thoughts were something like this, if i can transform first the original formula in a formula that is satiasfied only by all true assignament iff the original formula is satisfied, then do the same thing as before. So something like this: $$f(\phi(x_1,\dots,x_n))=\psi(x_1,\dots,x_n)\bigwedge_{i=1}^{n}y_i'$$ where $\psi(x_1,\dots,x_n)$ has only all true assignament iff $\phi(x_1,\dots,x_n)$ is satisfied. But i don't know how to construct $\psi$.