Is the shortcircuit behaviour of Python's any/all explicit?

Question

Prompted by the discussion here

The docs suggest some equivalent code for the behaviour of all and any

Should the behaviour of the equivalent code be conside

Answer 1

It HAS to short circuit, since it could be given an unbound iterable. If it did not short circuit then this would never terminate:

any(x == 10 for x in itertools.count())