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Logically, when I think about p unless q I want to say that it is equivalent to q -> ~p, but the only equivalence is ~q -> p. I verified by truth table that my intuition is wrong.

An example of why I want to think this way: I will go golfing tomorrow unless it rains in my mind is equivalent to If it rains tomorrow then I will not go golfing.

Is this basically a similar case to how when we state implications in general English that we imply the biconditional, even though it is a illogic thing to do?

How can I think about this when telling myself not to follow my intuition in this case? Is the reason that q -> ~p is wrong that p unless q doesn't say anything about what will happen if q is true?

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"I will go golfing tomorrow unless it rains" says what you will do tomorrow when it isn't raining. It doesn't actually say anything about what you will do when it is raining.

"If it rains tomorrow, then I will not go golfing" says what you will do when it is raining. It doesn't actually say anything about what you will do when it isn't raining.

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  • $\begingroup$ There is certainly a connotation, in the first quoted sentence, that I won't go golfing if it does rain. This is similar to the way that many disjunctions have a connotation of being exclusive disjunctions, even though when we formalize them in logic we treat them as inclusive disjunctions by conventions. Here, the most natural reading of "unless" in the first sentence is "I will go golfing tomorrow if and only if it does not rain", but by convention some do not read it that way for the purposes of logic. $\endgroup$
    – Carl Mummert
    Oct 14, 2015 at 11:33
  • $\begingroup$ While it may suggest other plans, it does not close the door on golfing in the rain. I.e., if it rained and I decided to golf anyway, that choice does not contradict the statement, like not golfing on a sunny day would. The second statement is just the opposite: golfing in the rain will contradict it, and not golfing in the sun does not contradict it. $\endgroup$
    – Paul Sinclair
    Oct 14, 2015 at 16:38
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    $\begingroup$ In normal English, it does contradict the statement, though - in normal English, if you say "I am going golfing unless it rains", you do mean "I am going golfing if and only if it does not rain". The convention in logic to treat the "unless" as something other than an "if and only if" is somehow out of sync with the natural language meaning of the statement. $\endgroup$
    – Carl Mummert
    Oct 14, 2015 at 17:00
  • $\begingroup$ I'll have to disagree with you on that. That has never been an interpretation I've had for "unless", and nor has it necessarily been an interpretation I've noted from others. $\endgroup$
    – Paul Sinclair
    Oct 14, 2015 at 22:04
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Regarding your golfing activity:

I will go golfing tomorow (without a doubt) unless it rains (then it is not sure that I will go golfing, but I may go).

Now the only thing you can extract from that is that if you did not go golfing, then it had to rain. (Or else you would have gone golfing).

So $p$ unless $q$ has to be understood as : $p$ is true is $q$ is false. But if $q$ is true, then maybe $p$ is true, maybe it is false.

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  • $\begingroup$ Consider another example: I will eat the last cookie, unless you already ate it. $\endgroup$
    – Carl Mummert
    Oct 15, 2015 at 0:26
  • $\begingroup$ This suggest that p has to be false, no? $\endgroup$
    – Jean-François Gagnon
    Oct 15, 2015 at 3:36
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An example of why I want to think this way: I will go golfing tomorrow unless it rains in my mind is equivalent to If it rains tomorrow then I will not go golfing.

I will go golfing tomorrow unless it rains says that you will go golfing if it does not rain.

If it rains tomorrow then I will not go golfing says nothing about what you will(or won't) do if it does not rain.

So, clearly they are not equivalent statements. They don't convey the same information. The actual equivalent statement is:

If it does not rain tomorrow then I will go golfing

"I will $P$ unless it $Q$" is equivalent to "If it not $Q$, then I will $P$", hence to: $\neg Q\to P$

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