# “At least one” — can this be translated in my problem? [closed]

Is the only way to replace "at least one" style questions just doing "total minus none"?

-
It likely depends on context. Do you have some samples you can add to your question? –  Amzoti Dec 30 '12 at 4:32
@Amzoti I have a grid of points and I can only travel in monotonic paths from one corner to the other, but there are arbitrary points here and there that I wish to avoid. So if I am to count the total number of "good" paths, I must do total paths - "bad" paths where a bad path = at least one bad point encountered, right? –  user51819 Dec 30 '12 at 4:34
I have trouble seeing how that has anything to do with "at least one". I.e. it sounds significantly more difficult to count the total good paths than to just prove there is a good path. –  peoplepower Dec 30 '12 at 4:36
Well I mean a bad path is a path for which "at least one" bad point is encountered, since you could theoretically go through many bad points. –  user51819 Dec 30 '12 at 4:36
OK. Put your example, in full detail, in the question. –  peoplepower Dec 30 '12 at 4:37

## closed as not a real question by Alexander Gruber♦, Ittay Weiss, Fabian, Hagen von Eitzen, Davide GiraudoDec 30 '12 at 11:29

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

At least one satisfies $\phi(x)$ is $\exists x \phi(x)$. The negation is none satisfies $\phi(x)$, which is $\forall x \lnot \phi(x)$

-
Sorry, what does this mean? –  user51819 Dec 30 '12 at 4:41
@user51819: I thought it was a direct answer. The first is how to write "at least one" as "there exists an $x$ such that $\phi (x)$ where $\phi (x)$ is a statement with one free variable. The second is how to say there are no $x$ such that $\phi(x)$. You can replace not the first by the second, which seems to me intuitive and is acceptable in logic. –  Ross Millikan Dec 30 '12 at 4:45

Since, a path is "bad" if it contains atleast one "bad" point. Total number of paths - Total number of bad paths = total # of paths which contain no bad point.

-