Any idea why all([]) is true while any([]) is false

>> all([])

ans =

     1

>> any([])

ans =

     0

[Ryan]

It's written into the documentation as such, but no explanation is given.

Answer 1

[Daniel Shub]

This post by Loren lead to some comments that address the issue, especially the one by Matt Fig.

Answer 2

[the cyclist]

I can't say I know definitively, but I expect that one reason is for consistency when taking the union of sets with the empty set. For example, one would want

all(union(true,[]))

to be true, and also

any(union(false,[]))

to be false. The definitions in your question make sense in that context.

[Khaled Hamed]

I have just noticed the same with NaN! more confusing

>> all(nan)

ans =

     1

>> any(nan)

ans =

     0