logic

  P->~P : ~P
philosophy boffins! anyone know how to prove this is a legit sequent without using RAA, or give hint to a method??? much appreciated =D

[math]P \Rightarrow \neg P[\math] is logically equivalent to
[math[\neg P \wedge \neg P = \neg P[\math]

There's a simple brute force proof.

P is either true or false.  Thus, P -> ~P is either
a) T -> F
b) F -> T

for
a) P is true, but statement is false
b) P is false, but statement is true

Thus, statement is equivalent to ~P