There was once a young squirrel by the name of Squirrely Squirrel the Squird.
This is Squirrely
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| Squirrely Squirrel the Squird holding a nut. |
Now, Squirrely loves nuts. Some say that Squirrely is never seen without a nut.
In fact, given P, a set of all pictures, S, a set of all pictures of Squirrely, and N, a set of all pictures with a nut, I can say that for all x in P such that for all y in P, x is in S if and only if y is in N. That is to say, for every single picture in the universe, that picture is a picture of Squirrely if and only if that picture has a nut. In fancy notation (for fancy people), that statement is written as ∀x∈P, ∀y∈P, x∈S ⇔ y∈P
Now, what prevented me from saying "for all x in S, x is in N"? ∀x∈S, x∈N
Or what about "for all x in P, if x is in S, then x is in N"? ∀x∈P, x∈S ⇒ x∈N
They're all equivalent statements... or are they?
Answer, these statements are not equivalent! My original statement claims that every picture of Squirrely has a nut AND every picture of a nut has Squirrely. The latter two statements only imply that Squirrely is always seen with a nut, but they say nothing about nuts being seen with Squirrely.
Yes, if there is a nut in the world, the nut will always be in the clutches of Squirrely.
Which brings us to our current point: logic is hard.
P.S. This is a cool website for symbols, but it keeps making a gray background when I copy and paste: http://en.wikipedia.org/wiki/List_of_logic_symbols
P.P.S. This is a link to a cool programmer who has no idea what he's doing, like me! I mean, we're both struggling to get through this course.
http://beardlessprogrammer.blogspot.ca/
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