Talk:Clavius's Law/Formulation 1/Proof 2

So as to cut down on all the necessary verbiage caused by your discomfort with $\vdash$, could be better expressed as:
 * $\left({\left({p \implies q) \implies p}\right) \implies p}\right) \vdash \left({\left({\neg p \implies p}\right) \implies p}\right)$

? --prime mover (talk) 08:06, 1 June 2013 (UTC)


 * There's nothing wrong with the current formulation. It's a model for how all such statements should be phrased. "If the sequent 'X' is valid, then so is the sequent 'Y'." is precisely what one wants to express. &mdash; Lord_Farin (talk) 09:43, 1 June 2013 (UTC)


 * Meh. Afraid the subtlety is beyond me. --prime mover (talk) 10:26, 1 June 2013 (UTC)