How to negate a relation in logic?

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This may be a dumb question but I am curious if this has a proof or if it's more of an axiom of logic and relations.

Consider the statement:

$$A = (x > 0)$$

Now we negate this:

$$\lnot A = (x \leq 0)$$

What is the logical reasoning behind this? It's intuitive here because we all know what these symbols mean without thinking too much, but what if it were some other symbol?

How do we know how to "negate" a relation?

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The real numbers are totally ordered, meaning exactly one of the following is true: $a < b$, $a = b$, or $b < a$. We write $b > a \iff a < b$ and $ a \leq b$ to mean $a < b$ or $a = b$.

From those facts, we can derive the fact as follows:

$$\left[ \left((a < b)\wedge\neg(a = b)\wedge\neg(a > b)\right)\\ \vee \left(\neg(a < b)\wedge(a = b)\wedge\neg(a > b)\right)\\ \vee \left(\neg(a < b)\wedge\neg(a = b)\wedge(a > b)\right)\right]\\ \wedge \neg(a > b) \\\implies (a < b) \vee (a = b) \iff (a \leq b)$$

Note the negations in each clause of the total order; these are logically necessary to exclude the possibility that another relation holds at the same time ("exactly one is true").

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There are $3$ possibilities for $x$:

  • $x<0$
  • $x=0$
  • $x>0$

If $x>0$ then the statement "$x<0$ or $x=0$" is false. The last statement is written as "$x\leq 0$".

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This is not logical; it depends on the particular meaning of the symbols. In fact, your example is not always true. Suppose the domain of $x$ is the complex numbers. Let $$A = (x > 0)$$ Then $$\neg A = \neg (x>0)$$ which is not equivalent to $$B = (x \leq 0)$$ Note that $\neg A$ is true of $x=i$ (since $A$ is false of $x=i$, as the $>$ relation holds only between real numbers), but $B$ is not (again because the $\leq$ relation holds only between real numbers).

So the equivalence between $\neg A$ and $B$ is true only in the domain of real numbers (or other similar domains), and depends on a property of real numbers, which is not a logical property.