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One of the exercises in Chiswell's Mathematical Logic is to prove the following sequent

$\{\neg(\phi\leftrightarrow\psi)\}\vdash((\neg\phi)\leftrightarrow\psi).$

I'm not interested in deriving this statement (which Chiswell says is "hard"), but giving it a natural interpretation. It's the type of statement which seems easy to apply, and yet, with a fallacious result. The following, for example, is manifestly false:

If it's not the case that rain is the same as precipitation, then, non-rain is the same as precipitation.

Here "is the same as" is a gloss for "iff-then".

What's my mistake here? And, what would be a correct, natural interpretation of this sequent?

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  • $\begingroup$ I think your mistake is thinking that in the real world "non-something" has only one possibility like when one negates a boolean function. $\endgroup$ – Javi Jul 3 '18 at 21:52
  • $\begingroup$ The result seems obviously false. Maybe I'm missing something. $\endgroup$ – Sarah Griffith Jul 3 '18 at 22:05
  • $\begingroup$ If $\phi$ and $\phi$ are each true or false, then $\lnot ( \phi \leftrightarrow \psi)$ says they do not have the same truth value. Hence $\lnot \phi$ and $\psi$ do have the same truth value. The $\leftrightarrow$ as a logical connective means "have the same truth value", not "are the same". However, isn't logic as much a part of the real world as anything else? $\endgroup$ – Carl Mummert Jul 3 '18 at 22:10
  • $\begingroup$ Basically, your "mistake" is that you are negating names (rain): you have to negate sentences : "If it's not the case that (5 is odd iff 5 is even), then, (5 is not odd) iff (5 is even)". $\endgroup$ – Mauro ALLEGRANZA Jul 4 '18 at 6:03
  • $\begingroup$ @MauroALLEGRANZA Then why can't we say that "If it's not the case that (it's raining outside iff it's precipitating outside), then, (it's not raining outside) iff (it's precipitating outside)"? This, after all, uses sentences, correct? $\endgroup$ – Doubt Jul 4 '18 at 18:21
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Rain and precipitation may be thought of as defined by unary predicates $R, P$ respectively true of precisely that which is rain and precipitation. To say that rain is not the same as precipitation essentially means $\neg \forall x (R(x) \leftrightarrow P(x))$. This is equivalent to $\exists x \neg (R(x) \leftrightarrow P(x))$, which by the result of the exercise implies $\exists x( \neg R(x) \leftrightarrow P(x))$. Glossing, if rain is not the same as precipitation, there's something which is not rain if and only if it is precipitation. (Snow, for instance, is such a thing.) That there are statements like $\exists x( \neg R(x) \leftrightarrow P(x))$ which don't gloss into very simple statements of ordinary language suggests the language of logical symbolism is sometimes more expressive than ordinary language.

To say that non-rain is the same as precipitation would be to say $\forall x( \neg R(x) \leftrightarrow P(x))$.

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Long comemnt

I think that the issue is about the correct reading of the notion of logical inference.

The formal definition of a valid logical inference is : every case where the premises are TRUE, also the conclusion must be TRUE.

Thus, in order to understand the "real world" counterpart of the above inference, we have to set up an example where the premise is TRUE.

But the premise is the negation of a biconditional; thus, the example must be based on a FALSE bi-conditional.

The issue is that in order to correctly evaluate the truth value of a sentence, we cannot change "state of affair".

What I mean, is that a typical sentence is "Napolean died in St.Helen"; this is TRUE now, was TRUE yesterday and will be TRUE tomorrow.

This is not the same with "it is raining", unless we specify : "it is raining here and now".

In that case, we have two possibilities: either it is the case that it is raining (here and now), in which case it is also the case that it is precipitating (here and now), and thus the bi-conditional :

it's not the case that (it's raining outside iff it's precipitating outside)

is FALSE, and the condition of valid inferential step is not applicable.

Either way, it is not the case that it is raining (here and now), but it is snowing, in which case it is again the case that it is precipitating (here and now), and thus the bi-conditional :

it's not the case that (it's raining outside iff it's precipitating outside)

is TRUE, and the condition of valid inferential step is applicable.

But in this case, we have that the conclusion :

"(it's not raining outside) iff (it's precipitating outside)"

is TRUE, because we have : TRUE iff TRUE.

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