Proving that $∀,\;(∀\:(≥⟹≥)⟹≥)$ Edit: My apologies, I'd framed my question wrongly, and I'm sorry for the confusion. What I was trying to ask is what Arther pointed out:
is there a way to prove that $$∀,\;(∀\:(≥⟹≥)⟹≥)?$$

Previous question: Given that $$(a \geq b) \implies (a \geq c),$$ where $a, b, c$ are all positive integers. I can't seem to rigorously prove that
$$b \ge c.$$ The result logically makes sense, since you can just draw a graph and visualize what the implication is saying. However, is there a formal way of mathematically inducing the second statement from the first? I've tried learning about entailments, but I don't see how that can apply here.
 A: 
is there a way to prove that $$∀,\;(∀\:(≥⟹≥)⟹≥)?$$

Proof: Since
\begin{align}
&∀,\;(∀\:(≥⟹≥)⟹≥)\\
\equiv{}&∀,\;\exists \;((≥⟹≥)⟹≥)\\
\equiv{}&∀,\;\exists \;((≥\quad\text{and}\quad <)\quad\text{or}\quad ≥)\\
\equiv{}&∀,\;\exists \;(b\le a<c\quad\text{or}\quad ≥),
\end{align}
it suffices to deduce the last statement above.
Let $b$ and $c$ be arbitrary real numbers and put $a=b.$ Then exactly one of $(≥)$ and $(b= a<c)$ is true.
Therefore, for each real $b$ and $c,$ there exists a real $a$ such that $$(b\le a<c\quad\text{or}\quad ≥),$$ as required.

Answer to the OP's previous question:

Given that $$(a \geq b) \implies (a \geq c),$$ where $a, b, c$ are all positive integers. I can't seem to rigorously prove that
$$b \ge c.$$

The counterexample $(a,b,c)=(4,2,3)$ shows that $$\exists a,b,c\in\mathbb N\;\Big((a\geq b{\implies} a\geq c)\quad\text{and}\quad b<c\Big),$$
that is, $$(a\geq b{\implies} a\geq c)\kern.6em\not\kern-.6em\implies b\ge c.$$
A: You can rewrite the implication as
$$(a<b)\lor(a\ge c)$$
which is not necessarily $c\le b$.
E.g. $(0\le1)\lor(0\ge2)$ is true but not $2\le1$.
