Tricky proof in Natural Deduction [¬∀x∃y¬Rxy ⊢ ∃x∀yRxy]-help appreciated

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As it says in the title, I'm trying to prove the statement ¬∀x∃y¬Rxy ⊢ ∃x∀yRxy in the system of Natural Deduction, with basic rules for negation intro and elimination, and existential or universal quantifier intro and elim.

I can use the premiss to create a contradiction, so the penultimate line might be:

¬∀x∃y¬Rxy ∀x∃y¬Rxy

     ∃x∀yRxy

But how do I derive the assumption ∀x∃y¬Rxy? It might be:

               ¬Rab 
             ∃y¬Ray              
           ∀x∃y¬Rxy 

So ¬Rab must be derived from some assumption Rab that leads to contradiction-that's where I'm stuck! Any help appreciated-this is not my homework or anything, it just bugs me!

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We need :

1) $¬∀x∃y¬Rxy$ --- premise

2) $¬∃x∀yRxy$ --- assumption [a] : the assumption to be contradicted, deriving (by Double Negation) the sought conclusion : $∃x∀yRxy$

3) $∀yRxy$ --- assumption [b] : in order to derive a contradiction with 2) to conclude, by $¬$-intro, with : $¬∀yRxy$, discharging [b].

Now we need a "detour", in order to derive from $¬∀yRxy$ the classically equivalent : $∃y¬Rxy$.

Having derived $∃y¬Rxy$, we can "generalize" it to $∀x∃y¬Rxy$. We can do it, because neither in the premise nor into assumption [a] $x$ is free.

Thus, $∀x∃y¬Rxy$ contradicts the premise 1) and we can conclude by Double Negation again with :

$∃x∀yRxy$,

discharging [a].