Webpage title: Are these logical equivalences truly equivalent?

[Bashyboy]

Say that I have p represent, "Swimming at the NJ shore is allowed;" and say that q represents, "Sharks have been spotted near the shore."

If I have the compound proposition $\neg p \rightarrow \neg q$, then it is equivalent to writing $q \rightarrow p$

Yet, that would suggest that the two propositions in English are equivalent. To be precise, "If swimming at the NJ shore is not allowed, then the sharks have not been spotted near the shore." and "If sharks have been spotted near the shore, then swimming at the NJ shore is allowed." Personally, I do not see how these to compound propositions, in English, are remotely equivalent.

 

[christoff]

Well, in English, those two propositions are silly. But they are still equivalent. I'm just going to alter the phrasing a bit to simplify the explanation...
p="I will swim"
q="There are sharks"

So ¬p→¬q means "If there are no sharks, then I will not swim".
and q→p means "If I will swim, then there are sharks".

"If there are no sharks, then I will not swim"... If I will swim (p), then there must be sharks. Why? Because "If there are no sharks, then I will not swim". Since I cannot both swim and not swim, there must be sharks. Therefore, "If I will swim, then there are sharks".

"If I swim, then there are sharks"... If there are no sharks (¬q), then I cannot possibly be swimming, since "If I swim, then there are sharks". Since there cannot be both sharks and no sharks, I must not be swimming. Therefore, "If there are no sharks, then I will not swim".

The equivalence follows from p∧(¬p) being a fallacy for propositions p.