Anyone want to give me a verbal example for why ∀x( P(x)->Q(x) ) is not equivalent to ( ∀xP(x) ) -> ∀x( Q(x) )
Replying to @ZachWeiner
P(x) = x draws comics. Q(x) = x is ugly. Since the premise is false, it is true that (∀xP(x)) -> ∀x(Q(x)). But since I'm handsome and drew a comic, it isn't true that ∀x(P(x)->Q(x)).
7:48 PM - 12 Jan 2019
0 replies
0 retweets
1 like
Loading seems to be taking a while.
Twitter may be over capacity or experiencing a momentary hiccup. Try again or visit Twitter Status for more information.