[ edited 11th april 2019]

Does the distinction between syntax and semantics imply that ( rigorously) the negation of P should not be read as " P is false"?

I'll try by the following comments to explain why the question is not off-topic from my point of view.

It can be argued that logical operators have a syntactic side and a semantic side.

From a syntactic point of view , the negation operator is simply ( if I am correct) a function from the set of wff to the set of wff. So the expression "~P" does not refer to the truth value of P, it only refers to the image of the proposition symbol P under the function " ~ " ( that is , the function : negation).

But from a semantic point of view, the negation operator is a function from {T,F} to {T,F}, namely the function { (T,F), (F,T) }.

So could one say that, at least from a semantic point of view, " ~ P " could be read as " P is false" ?

Or is this reading absolutely erroneous, as not taking into account the strict distinction between syntax and semantics?

Let me add an example.

Does the formula (P --> ~P) rigorouly mean that " P is false" ? I think it would be more correct to say that the formula only means " not-P". So there would be a difference between " not-P" and "P is false".

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    $\begingroup$ It means 'not $P$' $\endgroup$ – user251257 Apr 4 '19 at 20:03
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    $\begingroup$ Note that "not P" is true if and only if "P" is false, so they mean the same thing $\endgroup$ – MPW Apr 4 '19 at 20:18
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    $\begingroup$ This question is useless unless you give a source for the bizarre statement that $\lnot P$ should not be read as meaning $P$ is false. $\endgroup$ – Rob Arthan Apr 7 '19 at 2:10
  • $\begingroup$ The reading is legitimate in the context of classical logic; see bivalence. But there are other logics, see e.g. Intuitionsitic Logic with a specific semantics where the above reading is not appropriate. $\endgroup$ – Mauro ALLEGRANZA Apr 11 '19 at 13:11
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    $\begingroup$ "So could one say that, at least from a semantic point of view, "$\lnot P$" could be read as "$P$ is false" ?" In classical logic: YES. $\endgroup$ – Mauro ALLEGRANZA Apr 11 '19 at 13:12

If $P$ is false, then $\lnot P$ is true; if $P$ is true, then $\lnot P$ is false. It is the negation of $P$, whatever truth value $P$ might have, so perhaps that's where the confusion lies.

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    $\begingroup$ Sure, but when you assert $\neg P$ you are asserting that $\neg P$ is true, i.e., that $P$ is false. So I would not say it's incorrect to say "$\neg P$ means $P$ is false." $\endgroup$ – Jair Taylor Apr 4 '19 at 20:20
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    $\begingroup$ I suppose it's an example of the use-mention distinction. In the context of a truth table, you are discussing $\neg P$, not asserting it. You are just looking at the symbol $\neg P$ and evaluating its truth or falsity. If $P$ is true, then of course the statement "P is false" is false. $\endgroup$ – Jair Taylor Apr 4 '19 at 20:25
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    $\begingroup$ The statement "$P$ is false" is a valuation of $P$, @NoahSchweber. $\endgroup$ – Shaun Apr 11 '19 at 1:17
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    $\begingroup$ @Shaun Would you say that reading "~P" as " P is false" implies a confusion between syntax and semantics? $\endgroup$ – user654868 Apr 11 '19 at 11:13
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    $\begingroup$ Yes, @EleonoreSaintJames :) $\endgroup$ – Shaun Apr 11 '19 at 12:28

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