I must proof the "Trichotomy Law for Real Numbers":
- Prop. 1: let be $\Bbb{R}$ a complete ordered field, then $$\forall x,y \in \Bbb{R}(x=y \vee x < y \vee x > y)$$
- Proof 1: by definition of $ \Bbb{R} $, $\geq$ (or $\leq$) is total relation therefore $$\forall x,y \in \Bbb{R}(x \geq y \vee x \leq y) $$ but $x \geq y \leftrightarrow x=y \vee x >y$ then $$\forall x,y \in \Bbb{R}(x < y \vee x > y \vee x=y) $$
But (Prop. 2) $\forall x,y \in \Bbb{R}$ exactly one of the following holds:
- $x=y$
- $x<y$
- $x>y$
therefore $$\forall x,y \in \Bbb{R}(\mbox{ or } x=y \mbox{ or } x < y \mbox{ or } x > y)$$ where $(\mbox{ or } x=y \mbox{ or } x < y \mbox{ or } x > y)$ is "logically" CLIC...
How can I proof the Prop. 2? I thought proof the following:
- $x=y \leftrightarrow x \ngtr y \wedge x \nless y$
- $x <y \leftrightarrow x \neq y \wedge x \ngtr y$
- $x>y \leftrightarrow x \neq y \wedge x \nless y$
because CLIC is equivalente to eq_CLIC
Is it correct? Thanks in advance!
$\endgroup$1 Answer
$\begingroup$If you look at it like so:
- $x=y \leftrightarrow (x \le y \wedge y \le x)$
- $x<y \leftrightarrow x \le y \wedge \neg(x \le y \wedge y \le x)$
- $x>y \leftrightarrow y \le x \wedge \neg(x \le y \wedge y \le x)$
Combining any two will yield a $(x \le y \wedge y \le x)$ and a $\neg(x \le y \wedge y \le x)$
Also since $\le$ is a total relation, if $(x \le y \wedge y \le x)$ is false, one of the last two must be true. Hence at least one of three is true.
$\endgroup$ 2
