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Are these two statements considered negation of each other?:

  1. All fruits are sweet.
  2. Some fruits are not sweet.

I guess there is no relation between the two. I guess we can only say that statement 1 renders statement 2 "invalid" and vice versa. But nothing else can be said (in terms of "negation" and any other), am I right ?

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    $\begingroup$ I would say that the true negation of "All fruits are sweet" is "Not all fruits are sweet", though it happens to be easy to show (by quantifier negation) that "Not all fruits are sweet" is equivalent to "Some fruits are not sweet". $\endgroup$
    – Frank Seidl
    Commented Feb 4, 2021 at 16:12
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    $\begingroup$ Also, logicians typically reserve the word "invalid" for arguments which don't follow the correct rules of inference. Claims don't have validities; they already have truth values. $\endgroup$
    – Frank Seidl
    Commented Feb 4, 2021 at 16:16
  • $\begingroup$ Ohh so these two statements are indeed negation of each other? Then, what about (3) All fruits are not sweet? Is it a negation of (1) or (2) in the question? I feel it is indeed a negation of (2), in the same way, that (1) is a negation of (2)? $\endgroup$
    – Rnj
    Commented Feb 4, 2021 at 16:23
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    $\begingroup$ One could argue that they aren't truly negations of one another, because symbolically they differ by more than just a "not" symbol. (See Rob's answer for the symbolic representations.) However, they are at least equivalent to each other's negations. Claim 3 is not the negation of either, and also not equivalent to either's negation. This is because one can imagine a world in which claim 3 has the same truth value as either 1 or 2. $\endgroup$
    – Frank Seidl
    Commented Feb 4, 2021 at 16:33
  • $\begingroup$ So, claim (3) is neither equivalent to nor negation of either claim (1) or claim (2), right? $\endgroup$
    – Rnj
    Commented Feb 4, 2021 at 20:53

1 Answer 1

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If you formalise these two statements in first order logic, they read something like:

  1. $\forall f \cdot F(f) \to S(f)$

  2. $\exists f \cdot F(f) \land \lnot S(f)$

where $F(f)$ asserts that $f$ is a fruit and $S(f)$ asserts that $f$ is sweet. The logical negation of 1 is $\lnot\forall f \cdot F(f) \to S(f)$ which (in classical logic) is logically equivalent to 2. Likewise the logical negation of 2 is equivalent to 1.

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  • $\begingroup$ Should it be $\vee$ instead of $\wedge$? And if yes, I am wondering what it means words (and how it "sounds" equivalent to 1)? $\endgroup$
    – Rnj
    Commented Feb 4, 2021 at 16:47
  • $\begingroup$ $\land$ is right but I'd put $\lnot$ in the wrong place. I've fixed it. $\endgroup$
    – Rob Arthan
    Commented Feb 4, 2021 at 16:49

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