$a \lor b$ is equivalent to $\neg a \implies b$. If I'm trying to prove "i am human" $\lor$ "i am in Mars", it seems to me that it should make a difference which of the two propositions you pick as $a$ to prove the other one.

If I were to prove $a \lor b$, since it is true that "i am human" then i can say that "i am human" $\lor$ "i am in Mars" is true. I wouldn't symmetrically be able to take the same approach in trying to prove the disjunction by proving "i am in Mars". How then, can we symmetrically apply the same approach in proving $\neg a \implies b$? Shouldn't it make a difference which of the two propositions we take as $a$ and assume the negation of?

Something feels off and I'm trying my best to put it in words. Sorry if I'm failing to do so. Is this the difference between Classical and Intuitionistic Logic?

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  • $\begingroup$ It is the case that $a\lor b$ is not in general equivalent to $\neg a\to b$ in constructive/intuitionistic logic. Further, since this equivalence holding for all $a$ and $b$ is equivalent to excluded middle holding, you could say that this is (at a technical level) the difference between classical and intuitionistic logic. However, the concerns you bring up are not really the motivation and aren't really solved by constructivism/intuitionism. For example, it is constructively true that $a\lor b\iff\neg a\to b$ if $a$ is a decidable proposition, i.e. that we constructively know $a\lor\neg a$. $\endgroup$ – Derek Elkins left SE Sep 23 '19 at 4:22
  • $\begingroup$ It does not matter which proposition we take as $a$ because an implication is logically equivalent to its contraposition, which is what you end up with if you take the other disjunct for $a$. $\endgroup$ – mrp Sep 23 '19 at 8:54

If $a =$ "I am human" and $b =$ " I am in Mars", and you happen to be human, then $\neg a \implies b$ is true because a false statement implies anything. On the other hand, $\neg b \implies a$ is true because a true statement is implied by anything.

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