TITLE: Double negation elimination in constructive logic
QUESTION [15 upvotes]: How can I prove that the double negation elimination is not provable in constructive logic?
To clarify, double negation elimination is the following statement: 
$$\neg\neg q \rightarrow q$$

REPLY [3 votes]: A very simple and efficient tree proof method (i.e. a tableau method without signed formulae) for intuitionistic logic has been published by Bell, DeVidi and Solomon in "Logical Options: An Introduction to Classical and Alternative Logics" 
http://books.google.fr/books/about/Logical_Options.html?id=zUVYx-bTLgMC&redir_esc=y
Here is the tree counter-model for the formula $\neg \neg p \to p$, à la Bell $et{}~ al.$:
$$\underline{?(\neg \neg p \to p)}^{\surd}$$
          $$ ? p $$
          $$ \neg \neg p$$
          $$ \underline{? \neg p }^{\surd}$$
                $$ p $$
The tree show a Kripke counter-model where, in a locality expressed by a space between two horizontal lines, $p$ is not known to be true (i.e. is false in the locality), while $\neg \neg p$ is proved, i.e. is known to be true.  The symbol $?$ says that the formula is not known to be true, and it sticks the formula in the locality. The symbol $\surd$ says that the formula is deactivated. Every formula without $\surd$ or without $?$ can pass through any horizontal line (truth is persistent). For more details, see Bell et al.'s book.
This is very simple proof method to echo to the very nice explanation given by Zhen Lin. (I wish to thank Zhen Lin warmly for his post.)
Reading this page again, I must add that, contrarily to what Ben Millwood claimed, double negation elimination is not equivalent to the law of excluded middle, because if it is true that $(\neg A \lor A)$ implies $(\neg \neg A \to A)$ in intuitionistic logic, the converse is not intuitionistically provable. The fact that $(\neg \neg A \to A) \to (\neg A \lor A)$ is not an intuitionistic theorem contradicts the claim that double negation elimination is equivalent to the law of excluded middle.