I have a problem $\Pi_1$ that I want to show that is NP-hard. I know that I must find an NP-hard problem $\Pi_2$ and a polynomial time reduction $f()$ from instances of $\Pi_2$ to $\Pi_1$ such that $I_2$ is an Yes-instance of $\Pi_2$ iff $I_1=f(I_2)$ is an Yes-instance of $\Pi_1$.
What if I find a (constant sized) family of reductions $f_i()$ such that $I_2$ is an Yes-instance of $\Pi_2$ iff at least one $f_i(I_2)$ is an Yes-instance of $\Pi_1$? Is this enough? Is there a way of translating this one in the "classical" definition? How to formalize this?
I know that in the second situation I can say that I can't solve $\Pi_1$ in polynomial time unless P=NP, but I'm no sure that is equivalent of saying that $\Pi_1$ is NP-hard.