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Exactly how does it mean to negate a statement?

Is $\lnot P$ just the opposite of P, or is it anything but P ?

For example, consider the statement

  • None of the basketball players are blond.  $(P)$

Which is its negation?

  • All the basketball players are blond.  (P's exact opposite)
  • At least one of the basketball players is blond.  (anything but P)

A side question: what is the precedence of ∧ (and), ∨ (or), ¬ (negation), $→$ (implies), etc.?

ryang
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Temp Jon
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  • The situations in which $\lnot P$ is true are precisely the situations in which $P$ is not true. – Karl Jan 12 '25 at 21:06

4 Answers4

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I think for your first question the best way to think about is $\neg P$ is "it is not the case that $P$." So in your example, if $P$ is "None of the basketball players are blond," then $\neg P$ is "it is not the case that none of the basketball players are blond" which is like saying "there is some basketball player with blond hair."

Nathan
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One way to keep these two apart in your mind is to remember that $p$ and $\neg p$ (1) cannot both be true at the same time, and (2) cannot both be false at the same time (one of the two must be true in a given situation). Considering this, it is not possible to simultaneously have "all blond" and "none blond" (unless there are no players), but it is possible to have neither "all blond" nor "none blond" situations being the case, for example if there are some blond and others not blond. By contrast, "none blond" and "some blond" can not happen at the same time, and also one of the two has to be true (either there is a blond or there isn't). Thus "some blond" is the correct negation of "none blond".

Regarding precedence, usually $\neg$ has the highest precedence and $\lor,\land,\to$ compete for a lower predecence - I would not recommend relying on the precedence order of these and just bracket everything. But I think the most common convention puts $\land$ above $\lor$ above $\to$, so that $A\land B\lor C\to \neg D$ is $((A\land B)\lor C)\to (\neg D)$.

Mario Carneiro
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  • I think I might be getting it. For example, if p = "You are always early", what would ~p be, other than "You are not always early" (not including 'not' in sentence) – Temp Jon Feb 04 '15 at 03:47
  • @TempJon Well, presumably you know what "not always early" means, that is indeed the correct negation here. But if you want to avoid the word "not" here, you might also say "you are sometimes late". – Mario Carneiro Feb 04 '15 at 05:35
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Is $\lnot P$ just the opposite of P, or is it anything but P ?

Negation:
 Statements $\,\lnot P$ (“not $P$”) and $P\,$ have opposite truth values regardless of the context.

Complement:
 Sets $\,A^\complement$ (“non-$A$”) and $A\,$ each comprise all the elements that are absent from the other.

  • None of the basketball players are blond  $(P)$

Which is its negation?

  • All the basketball players are blond  (P's exact opposite)
  • At least one of the basketball players is blond  (anything but P)

Shouldn't "anything but statement $P$" refer to every statement that isn't $P$ ?

As for the notion "the exact opposite of statement $P$", how about All the basketball players are not blond instead/too? (Unfortunately, this statement is actually equivalent to $P.)$

Now, by adding

  • At least one of the basketball players is not blond

to your three statements, we obtain the four canonical categorical propositions (no pair of which is equivalent)

  1. Some K is L
  2. Some K is not L
  3. Every K is L  (i.e., No K is not L)
  4. No K is L  (i.e., Every K is not L).

Observe that it is not the case that No K is L precisely means that Some K is L; so, propositions (1) and (4) negate each other; similarly, propositions (2) and (3) negate each other.

Note that merely switching 'some K', 'every K' and 'no K' does not generally negate a categorical proposition; for instance, the negation of Every K is L is

  • neither No K is L (since $\forall x\,(K(x)\to L(x))\;\leftrightarrow\;\lnot\exists x\,(K(x)\land L(x))\tag*{}$ is true when $K(x)$ means '$x ≠ x$')
  • nor Some K is L (since $\forall x\,(K(x)\to L(x))\;\leftrightarrow\; \exists x\,(K(x)\land L(x))\tag*{}$ is true when $K(x)$ and $L(x)$ both mean '$x = x$').

Hence, the negation of statement $P$ (None of the basketball players are blond) is At least one of the basketball players is blond but not All the basketball players are blond.

A side question: what is the precedence of ∧ (and), ∨ (or), ¬ (negation), $→$ (implies), etc.?

The precedence convention in first-order logic.

ryang
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Is $\lnot p,$ the negation of a statement $p,$ just the opposite of $p,$ or is it anything but $p\;?$

It is anything but $p$. So in your example, "At least one of the basketball players is blond" .

A side question: what is the precedence of ∧ (and), ∨ (or), ¬ (negation), $→$ (implies), etc.?

Depends on the definition; usually, it is $\neg > \land > \lor > \rightarrow > \leftrightarrow$.

Natalie Clarius
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