For instance, see Generalized Riemann Hypothesis. It conjectures that if $L(\chi, s) = 0$, and $0 \leq \Re(s)\leq 1$, then $\Re(s) = 1/2$. Then is there a function $f(s)$ that you can think of that satisfies that condition, i.e. $f(s) = 0 \implies \dots \ \Re(s) = 1/2$?
I thought of $f(s) = Re(s) - 1/2 = s^* + s + 1$, then it satisfies the weaker statement: if $f(s) = 0 \implies \Re(s) = 1/2$.
But this example alone is obviously uninteresting. Can we come up with more, preferably more interesting ones without going full circle back to the conjectured $L$-functions?
Firstly, we should restate RH as $L(\chi, s) = 0 \implies \Re(s) \in (-\infty, 0) \cup \{1/2\} \cup (1, \infty)$. Maybe that will help.
Obviously two functions satisfying the above condition, when multiplied together, yield another such function. Proof: $f(s)g(s) = 0 \implies g(s) = 0 \vee f(s) = 0 \implies \dots \ $ Thus such functions all together (including $L$-functions) already form a semigroup with identity as $1 \neq 0$ and sats the statement vacuously.
But we're not used to working with just semigroups so let's see if we can get a ring out of it.
Consider all functions $f: \Bbb{C} \to \Bbb{C}$, such that $f(s) \gt 0$ on all $\Re(s) \in [0,1/2) \cup (1/2, 1]$. Then you can add such functions, now we have a semiring. If we include the zero function then that's the only function that doesn't sat the RH statement. Alternative constructions include letting $f(s) \lt 0$ or $f(s) \gt 0$ on two interval halves of $[0,1]$.