What is the logic proof that ¬ (∃x)(Fx) => (∀x)(Fx → Gx)

Question

I need to prove the proposition ¬ (∃x)(Fx) => (∀x)(Fx → Gx) but haven't really found a good way. I started off this way:

1. ¬ (∃x)(Fx) Premise
2. ¬ (∀x)(Fx → Gx) Conclution
3. (∃x)[ ¬ (Fx → Gx)] Negation distribution in 2
4. (∃x)[¬(¬Fx˅Gx)] Material implication in 3

Feel like I'm close to the answer but still no concrete way to prove this. Any comments or suggestions would be really appreciated.

Answer 1

Before attempting to construct a formal proof of anything, you should have an informal reason to believe it is true. I suggest:

• $\lnot \exists \langle x,~ Fx \rangle$ means $F$ is always false
• $\forall \langle x,~ Fx \implies Gx \rangle$ follows from "false always implies anything ($G$ in this case)"

A constructed proof should follow that general outline. Here is a natural deduction one:

$$\begin{array} {l} % \quad \begin{array} {|l} \lnot \exists \langle x,~ Fx \rangle \\ \hline % \quad \begin{array} {|l} Fy \\ \hline % \exists \langle x,~ Fx \rangle \\ % \bot \\ % Gy \end{array} \\ % \forall \langle x,~ Fx \implies Gx \rangle \end{array} \\ % \lnot \exists \langle x,~ Fx \rangle \implies \forall \langle x,~ Fx \implies Gx \rangle \end{array}$$