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

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>> all([])
ans =
1
>> any([])
ans =
0
  2 Comments
Khaled Hamed
Khaled Hamed on 29 Jul 2012
I have just noticed the same with Nan! more confusing
>> all(nan)
ans =
1
>> any(nan)
ans =
0

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Accepted Answer

Daniel Shub
Daniel Shub on 29 Jul 2012
This post by Loren lead to some comments that address the issue, especially the one by Matt Fig.

More Answers (1)

the cyclist
the cyclist on 29 Jul 2012
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.
  1 Comment
Khaled Hamed
Khaled Hamed on 29 Jul 2012
I have just noticed the same with NaN! more confusing
>> all(nan)
ans =
1
>> any(nan)
ans =
0

Sign in to comment.

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