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I need to prove:
$\forall x.p(x) \Rightarrow \neg \exists x.\neg p(x)$
I know I probably have to assume $\exists x.\neg p(x)$ to reach a contradiction. However, I am not quite sure about how to reach the contradiction.
The premise is that $\forall x.p(x)$, so maybe I had to prove $\exists x.\neg p(x) \Rightarrow \neg \forall x.p(x)$ in this case. However, to get $\neg \forall x.p(x)$ required assuming $\forall x.p(x)$ in advance, which is the same as the premise. Is it feasible? Whether such scenario would cause a endless loop?
I am not sure how could I get a contradiction for this case, any help would be appreciated, thanks!

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    Shouldn't this follow from De Morgan's laws. In which you can prove it directly with a logical map. in either case, you are trying to prove that you can use contrapositive logic to prove things are true. – 2017-02-25
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    Yeah. it's quite intuitive, what I want here is a systematic structural proof. – 2017-02-25

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Different systems define the Existential Elimination (or Existential Instantiation) differently, but here is a proof in the Fitch system I use:

enter image description here

EDIT

OK, so your system is indeed a good bit different from the one I use above. Here is a proof in your system:

  1. $\forall x P(x)$ Premise

  2. $\quad \exists x \neg P(x)$ Assumption

  3. $\quad \quad \neg P(x)$ Assumption

  4. $\quad \quad \quad \forall x P(x)$ Reiteration 1

  5. $\quad \quad \quad \neg P(x)$ Reiteration 3

  6. $\quad \quad \forall x P(x) \rightarrow \neg P(x)$ $\rightarrow \: I$ 4-5

  7. $\quad \quad \quad \forall x P(x)$ Assumption

  8. $\quad \quad \quad P(x)$ $\forall \: E$ 7

  9. $\quad \quad \forall x P(x) \rightarrow P(x)$ $\rightarrow \: I$ 7-8

  10. $\quad \quad \neg \forall x P(x)$ $\neg \: I$ 6,9

  11. $\quad \neg P(x) \rightarrow \neg \forall x P(x)$ $\rightarrow \: I$ 3-10

  12. $\quad \forall x (\neg P(x) \rightarrow \neg \forall x P(x))$ $\forall \: I$ 11

  13. $\quad \neg \forall x P(x)$ $\exists \: E$ 2,12

  14. $\exists x \neg P(x) \rightarrow \neg \forall x P(x)$ $\rightarrow \: I$ 2-13

  15. $\quad \exists x \neg P(x)$ Assumption

  16. $\forall x P(x)$ Reiteration 1

  17. $\exists x \neg P(x) \rightarrow \forall x P(x)$ $\rightarrow \: I$ 15-16

  18. $\neg \exists x \neg P(x)$ $\neg \: I$ 14,17

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    I'm sorry that I'm not familiar with your system's notation. Is step 3 equivalent to $\forall a \neg P(a)$? – 2017-02-25
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    @Han Yeah, I was afraid your system might be different ... How is your Existential Elimination rule defined? ... And no, you don't want to use $\forall a \neg P(a)$, since $a$ is a constant, and you can only quantify over variables. – 2017-02-25
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    EE is such, $\exists x.p(X)$ $\forall x.(p(X) \Rightarrow q(X))$, then we could obtain $q(X)$. So the figure you show is a workflow without details, right? – 2017-02-25
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    @Han. I really doubt that that is how EE is defined ... i have seen many different particular implemenations of EE, but I have never seen anyone that requires you to have a $\forall$. indeed, how would you infer $\exists x (P(x) \land Q(x))$ from $\exists x (Q(x) \land P(x))$ using this as the EE rule? Also, no, the figure I show is not just a workflow ... It does have all the details for the particular proof system I work with. – 2017-02-25
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    Sorry for the late response. This EE is defined under http://logic.stanford.edu/intrologic/notes/chapter_08.html. This definition seemed like Or Elimination. The question you proposed is solvable in this case, s.t. assume $Q(x)\wedge P(x)$ then using EI to obtain $Q(x)\wedge P(x) \Rightarrow \exists x.(P(x)\wedge Q(x))$. Then using UI to introduce the $\forall$ symbol. Last step is a direct application of EE. – 2017-02-26
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    @Han Huh! I must say, I have never seen such an implementation of EE! But you're right, it does indeed require a universal! Well, OK, no worries, my proof can easily be modified to fit your format.: On line 6, use $\forall \: I$ On the subproof to get $\forall x (\neg P(x) \rightarrow \bot)$. Then on line 7, use EE on lines 2 and 6 to get $\bot$. And line 8 will be like the original line 7. – 2017-02-26
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    @Han Oh, wait, their universal Introduction rule is different as well from how I use it. ... OK, let me just edit my Answer. Give me 10 minutes or so... – 2017-02-26
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    Thank your for your kindly help these days. I understood all your steps and I thought your solution is reasonable enough, however, because of the difference of Fitch System, I am still struggling with this problem... – 2017-02-26
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    @Han Do you have the contradiction ($\bot$)? And can you maybe send me a link to the propositional logic rules? – 2017-02-26
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    We do not have the contradiction symbol, yeah, that's the main part that I was stuck. I think most of the rules can be checked with this url: http://logic.stanford.edu/intrologic/glossary/glossary.html. It seemed that rules about $\forall$ and $\exists$ were excluded, both of them can be seem from here: http://logic.stanford.edu/intrologic/notes/chapter_08.html – 2017-02-26
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    @Han Oh boy! They're not making this easy, do they?! Anyway, let me modify my proof yet again! – 2017-02-26
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    @Bram28, yeah, this might be the hardest one for this problem set. I don't know why the discussion forum was removed, and I was stuck almost a week or so... – 2017-02-26
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    @Bram28: You may be interested to know that it is easiest to use such an existential elimination rule to represent Fitch-style natural deduction in a sequent calculus like in [this post](http://math.stackexchange.com/a/1822074/21820). I also found a very very brief summary of many ND systems at http://ualberta.ca/~francisp/papers/pelletierNDtexts.pdf – 2017-02-26
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    @Han Holy cow! I got it, but that was a major pain!! What a sadistic system of rules. :( That EE rule requiring you to have a universal makes for quite roundabout proofs, and the same goes for not having an explicit $\bot$ and associated rules. Look how much easier this is in the system I use.... Oh well! – 2017-02-26
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    @user21820 Hey, thanks for that! Especially that paper is of great interest to me! I'll need some time to learn the formatting with those lines ... – 2017-02-26
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    @Bram28: Brilliant! Maybe this sadistic system aims at training my thinking ability...BTW, the order of your steps 3&4 is fantastic!!! I always put step 4 first and end up with endless loop...My reputation is less than 15 so I cannot give you a agreement. ::>_<:: – 2017-02-26
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    You're welcome! =) By the way, LaTeX formatting of those lines can be greatly helped by using a macro. Try `$\def\block#1#2{\begin{array}{|l}#1 \\ \hline #2\end{array}}$` or `$\def\block#1#2{\begin{array}{|l}\!\underline{\ #1\ } \\ #2\end{array}}$`, either of which would allow you to then write just `$\block{A}{\block{\neg A}{\bot} \\ \neg \neg A}$` to get $\def\block#1#2{\begin{array}{|l}#1 \\ \hline #2\end{array}} \def\bloc#1#2{\begin{array}{|l}\!\underline{\ #1\ } \\ #2\end{array}}$ $$\block{A}{\block{\neg A}{\bot} \\ \neg \neg A}$$ or $$\bloc{A}{\block{\neg A}{\bot} \\ \neg \neg A}$$ =D – 2017-02-26
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    @Han Generally, I think these logic systems will make your thinking more organized ... but there are some proofs where I have my doubts what exactly the purpose is ... This is one of those cases! Anyway, thanks for the challenge :) – 2017-02-26
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    @user21820 Oooh, very nice! Thanks!! – 2017-02-26
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I may be misunderstanding the question, but it seems like the contradiction follows almost directly from supposing the negation of the conclusion. By negating the conclusion, we have $\exists x$ such that $\neg p(x)$, call it $x_0$. So, by using that particular $x_0$ we have a contradiction to the premise that $p(x)$ for any $x$.

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    Yeah, this problem is really easy because it's intuitive. However, here I need a systematic structure. – 2017-02-25