How can I prove this statement without using reduction to absurdity?

Question

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$\forall a,b\in\mathbb R[\forall c\in \mathbb R(c>a\implies c>b)\implies a\ge b]$

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Answer 1

This seems impossible, because the statement is false for non total orders in general. For example, take the set $S=\{x,y,z\}$ with the order $x>y$ and $x>z$, but $y$ and $z$ are not related. Then we have two elements ($x$ and $y$) that meet the condition $\forall c\in S(c>z\implies c>y)$ but $\neg(z\ge y)$.

Thus, we must use that the order is total, that is, we need to show that $$\forall a,b\in\Bbb R[(\forall c\in\Bbb R(c>a\implies c>b))\implies \neg(a <b)$$ and the only way to introduce negations is via reductio ad absurdum.

I've said that this seems impossible, but I don't mean I have proved it. There are non total orders that meet the statement, for example the divisibility in $\Bbb Z$, so perhaps there is still a way.

Answer 2

We may use the geometric intuition and the concept of inf in $\mathbb{R}$ to obtain a direct proof of your statement, i.e., without using reduction to absurdity.
Detailed proof. Fix $a,b\in \mathbb{R}$.
$\{c\in \mathbb{R}|c>a\}\subset \{c\in \mathbb{R}|c>b\}$ (by hypothesis and the Venn diagram)
$b=\inf \{c\in \mathbb{R}|c>b\}$ is a lower bound of $\{c\in \mathbb{R}|c>a\}$. Consequently, $a=\inf \{c\in \mathbb{R}|c>a\}\ge b$. Q.E.D.
For detailed discussions about reduction to absurdity, read (1). Example 6.169 [especially, the last two sentences in part IV] of https://sites.google.com/view/lcwangpress/%E9%A6%96%E9%A0%81/papers/mathematical-methods and
(2). https://sites.google.com/view/lcwangpress/%E9%A6%96%E9%A0%81/papers/reduction-to-absurdity.