Talk:Peirce's law

interesting proof. ( my basic logic says "if p is false then (p->q) is true and so p is true and so p must be true" ).<Hillgentleman| ~ | 書> 04:15, 31 March 2010 (UTC)[reply]

I think a more exact paraphrase would be:

"If $(p\Rightarrow q)$ implies $p\!$ then $p\!$ must be true"

One of the things of interest here is the fact you can start with a very weak ("intuitionistic") axiom system for logic, one where you don't even have the Law of Excluded Middle (LEM) yet, and adding Peirce's formula as an axiom is equivalent to having LEM. Jon Awbrey 04:26, 31 March 2010 (UTC)[reply]