I know that this definition is unrelated to the law of the excluded middle, but as a beginner in logic (I've studied the first half of Chiswell and Hodges' Mathematical Logic), the use of the name 'negation', and the fact that in truth tables the negation of a true proposition is false, and vice-versa, make me think of A $\rightarrow \bot$ as "A implies absurdity, therefore the opposite of A is true". But isn't that the law of the excluded middle, or at least a very similar idea?
1 Answer
This is a way of defining negation ($\lnot$), assuming conditional ($\rightarrow$).
When $A$ is true, $A \rightarrow \bot$ will be $TRUE \rightarrow FALSE$, which is false, and thus it works.
Of course, in order to define $\lnot$, we have to assume a new "primitive" concept : the falsum or absurdity ($\bot$).
Usually, in natural deduction $\bot$ is primitive; with it the basic rules for minimal and intuitionistic logic are stated.
In order to "extend" the set of rules to cover classical logic, we have to add a rule; it can be one of : RAA, Exclude Middle, Double Negation or Dilemma (see this post).
Added
See Dirk van Dalen, Logic and Structure (5th ed - 2013), page 29-on.
The connectives are usually "managed" by a couple or rules : introduction and elimination.
Negation is defined from $\bot$ and the rules for $\bot$ in classical logic are :
$$\frac {\bot} A \, (\bot \text {-E})$$
also called : ex falso quodlibet; and :
$$\frac {\frac {[\lnot A]} \bot } A \, \text{RAA}$$
Only with (RAA) we can derive LEM, i.e. $A \lor \lnot A$.
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$\begingroup$ I've seen the post you linked, thanks. I get it that it works, that is, when A is true, $\lnot$A is false, and vice-versa. But then doesn't it mean that ($A \lor \lnot A$), the LEM, is always true? $\endgroup$– mcostaCommented May 18, 2014 at 17:19
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$\begingroup$ Yes, but only in classical logic; in intuitionistic one, this is not true. It depend on the rules you use. (see my edit of the answer). $\endgroup$ Commented May 18, 2014 at 17:45