Proving that $∀,\;(∀\:(≥⟹≥)⟹≥)$

Question

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 $$∀,\;(∀\:(≥⟹≥)⟹≥)?$$

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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.

Answer 1

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.

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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, $$(a\geq b{\implies} a\geq c)\kern.6em\not\kern-.6em\implies b\ge c.$$

Answer 2

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 $2\le1$.