Why do people say RAA(Reductio Ad Absurdum) is the same as $(\bot E)$? $(\bot E)$ is > $\bot\vdash\psi$. RAA(Reductio Ad Absurdum) says > If $\\{\Gamma,\neg\psi\\}\vdash\bot$, then $\\{\Gamma\\}\vdash\psi$. Yet, one of the solutions to my textbook exercises uses $(\bot E)$ and labels it (RAA) in its natural deduction proof as in the picture below. Is it a mistake? ![enter image description here]( **Update 1** : Is the picture below a correct usage of RAA in a natural deduction proof? ![enter image description here](

See:

* Dirk van Dalen, Logic and Structure (5th ed - 2013), page 30 and page 156.



Where the presentation consider both _intuitionistic_ and _classical_ logics, the distinction between:

> ($\bot$E): $\dfrac \bot \varphi$

and:

> (RAA): $\dfrac { [\lnot \varphi] \ldots \bot } \varphi$

has to be maintained.

* * *

You can see the answers to what-is-the-correct-reading-of-bot for an overview of the different rules fo the _negation_.