F*教程的第9章有一个例子：
b ::= x | true | false
e ::= b | let x = e1 in e2 | assert b | if b then e1 else e2
WP b P = P b
WP (let x = e1 in e2) P = WP e1 (fun x -> WP e2 P)
WP (assert b) P = b /\ P b
WP (if b then e1 else e2) P = (b ==> WP e1 P) /\ ((not b) ==> WP e2 P)
有人可以在这里解释这个符号吗？我知道这WP是前置条件和P后置条件，但什么是P b？应用于b?的后置条件
来自 IRC 频道#fstar 的讨论：
20:33:49 - {AS}: let us say we have a program (let abs x = if x >= 0 then x else -x)
20:33:58 - ollehar1: ok
20:34:02 - {AS}: and we want to show that (fun res -> res >= 0)
20:34:15 - {AS}: so we do
20:34:26 - {AS}: WP (if x >= 0 then x else -x) (fun res -> res >= 0)
20:34:28 - {AS}: right?
20:34:32 - ollehar1: ok
20:34:36 - {AS}: then we follow the calculations in the example
20:34:42 - {AS}: so we apply the if-rule and get
20:35:39 - {AS}: ((x >= 0) ==> WP x (fun res -> res >= 0)) /\ ((not b) ==> WP (- x) (fun res -> res >= 0))
20:35:55 - {AS}: here ==> is implication and /\ is and
20:36:05 - ollehar1: yes
20:36:07 - {AS}: conjunction
20:36:17 - {AS}: so we carry on
20:36:17 - ollehar1: but "not b" should be "not x >= 0"
20:36:24 - {AS}: Yes, sorry :)
20:36:39 - ollehar1: np
20:36:43 - {AS}: not x is a basic value
20:36:46 - {AS}: now*
20:37:04 - {AS}: so WP x (fun res -> res >= 0) reduces to x >= 0 according to the rules
20:37:14 - {AS}: and similarly with - x in the other side
20:37:31 - {AS}: (they don't really have minus, so I just used the common intuition for that rule)
20:38:05 - {AS}: so we end up with having ((x >= 0) ==> (x >= 0)) /\ (not (x >= 0) ==> (- x >= 0))
20:38:35 - {AS}: of course, if we simplify a bit the first conjunct is trivially true
20:39:00 - {AS}: and if you have an integer arithmetic solver, you find out that the second is also true
20:39:13 - {AS}: so the weakest precondition ends up being true