Suppose is true, and that .Originally Posted by SkyWatcher
Let be the subset of for which is false.
Then by the axiom there is a smallest element of ,
and by assumption.
Then is true and is false, which is
a contradiction of the assumption that .
Hence there is no such element and so there is no such
set . And so is true for every .
(Note here we are taking if we want to
use , we would have to start with
being true rather than in our base case.)