Proving a Derived Rule with Natural Deduction: Need Help!

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Hello smart people!
I was having some troubles proving this derived rule using Natural deduction:

¬(∃y.Q(y) ∧ T(y))
------------------------
∀x.Q(x) → ¬ T(x)

I got stuck in the very first line, because of the "NOT". I can't do anything if I don't take it out of there...

I know that ¬(a ∧ b) = ¬a ∨ ¬b, but I don't know how to prove that as well...

If I could only get past this very first step I'd be able to finish it, but I've been trying for hours and I can't get around it.
Please help!
 
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To prove DeMorgan's Law, you might want to write out a truth table.
 
It's also helpful to note that:

p \implies q \equiv \neg p \vee q
 
I would use the identity

##\exists x \phi\equiv \neg\forall x\neg\phi## together with the theorem ##\neg\neg\phi\leftrightarrow\phi## and then de Morgan's Law.

If you need more help you will need to specify which Natural Deduction system you are using. There are many. To specify yours you need to list all the rules of inference and replacement rules you have available to you.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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