# equivalence of implications

I've got a little trouble with logic. I'm reading a book about mathematics, and as an example for implication there is:

$$x > 3 \implies x>0$$

So, if $x$ is bigger than $3$, it implies that it is bigger than zero, okay. But in general, we say: $$a \implies b = \overline b \implies \overline a$$ and $$\overline{x > 3} = x \underline < 3$$ so if $x$ is not bigger than $3$, it is equals to or less than $3$. all fine here. But this means also that:

$$x > 3 \implies x>0 = x \underline < 0 \implies x \underline < 3$$

Okay, but I read it as: if $x$ is smaller or equals to $0$, it is smaller or equals to $3$. But if it is already smaller than $3$, how can it be equals $3?$

Seems not very logic to me :)

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The implication $x\le 0\implies x\le 3$ is still true because $x\le 0\implies x<3\implies x\le 3$ – Stefan Hamcke Aug 22 '13 at 18:42
Alraxite has a fairly good answer here. The last statement says that x it is the case that EITHER x is less than or equal to 3. It does not say which one of those possibilities holds, just that one of them does. – Doug Spoonwood Aug 23 '13 at 3:03

$x \leq 0 \implies x \leq 3$ is a true statement.
$x\leq 3$ means $x\lt3$ or $x=3$. So, as long as $x<3$ remains true, $x\leq 3$ also remains true. In the same way, $1\leq1$ or $2\leq 3$ are also true statements.
In general, $P\implies P \vee Q$ is a tautology.