Example One

$\displaystyle \langle \; \rangle \; x := 2*y \; \langle x = 2y \rangle $

(where $ x = \mathbb{R}$x$ \mathbb{R}$) is totally correct, since if we replace $ x$ by $ E = 2y$ in $ \alpha = \lq\lq (x=2y)''$ gives ``$ (2y = 2y)$'' which is just True

David Goodwin 2008-09-20