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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.

ryang
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ryanyxw
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    Take a= b. Then it is true that $a\ge b$ so that $a\ge c$. But since a= b, $b\ge c$ thus $c\le b$. – George Ivey Feb 16 '23 at 18:22
  • The problem statement is missing a few quantifiers. Are $b$ and $c$ fixed, and the given implication true for all $a$? Or are they all fixed, and the given implication just true for the three given numbers? Or what's going on here? – Arthur Feb 16 '23 at 18:37
  • @Arthur I reckon that if an implication is missing quantification, then it is implicitly universally quantified. – ryang Feb 16 '23 at 19:08
  • @ryang But there are plenty of ways to universally quantify. For instance, $$\forall a,b,c:(a\geq b\implies a\geq c)\implies b\geq c$$ isn't true. On the other hand, $$\forall b,c: (\forall a: a\geq b\implies a\geq c)\implies b\geq c$$is. Maybe some other combination might be true as well. It's impossible to tell which one is actually meant. Hence my question. – Arthur Feb 16 '23 at 19:28
  • Ok, write it this way then: $$\forall a,b,c((a\geq b\implies a\geq c)\implies b\geq c) $$ which isn't true, versus $$\forall b,c((\forall a(a\geq b\implies a\geq c))\implies b\geq c)$$which is. – Arthur Feb 16 '23 at 19:31
  • @Arthur To be clear, I’d call an open formula “implicitly quantified” iff its outermost quantifiers have been dropped. – ryang Feb 17 '23 at 07:50
  • @ryang You can still get into interpretation territory. Is $a\geq b\implies a\geq c$ a separate formula, with free $b,c$ and an implicitly quantified $a$, and then the whole statement a formula with implicitly quantified $b,c$, or is the entire thing a single formula with implicitly quantified $a,b,c$? It is fundamentally ambiguous, no matter how much you like implicit quantifiers. – Arthur Feb 17 '23 at 08:13
  • @ryang And I am specifically waiting to answer until we can clear up what exactly the OP actually wants to ask. Many problems like these are unfortunately posed, and this one has gone through an additional layer of translation, from the original author through the OP, to us. Given that they are likely asked by their original exercise to prove $b\geq c$, doesn't that mean we should at least try to clarify before we rush out to say the problem is wrong? – Arthur Feb 17 '23 at 08:51
  • "you can just draw a graph": my bet is that you didn't draw such a graph, which is 3D and would show the falsity of the statement. –  Feb 17 '23 at 09:00

2 Answers2

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

ryang
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  • This answer makes some assumptions about the order and scope of the implicit quantifiers involved (see comment thread on the question post). It may not be the answer that ryanyxw is after. – Arthur Feb 16 '23 at 21:04
  • Hello @Arthur, Yves's and my previous answers do, without making any unwarranted assumption, correspond to the OP's previous question. $\quad$ In any case, the OP has subsequently added the more interesting question raised by you, so I've expanded my post to add the new answer. – ryang Feb 28 '23 at 09:20
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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 $2\le1$.

  • This answer makes some assumptions about the order and scope of the implicit quantifiers involved (see comment thread on the question post). It may not be the answer that ryanyxw is after. – Arthur Feb 17 '23 at 08:18
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    @Arthur: you seem to read information that it not in the problem statement. A counterexample is enough. –  Feb 17 '23 at 08:35
  • I think the question is lacking information, you think it has all the information we need. We can argue about whether it's actually there, but at least be honest about who thinks it is. – Arthur Feb 17 '23 at 08:54
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    @Arthur: sorry, but I think that you are trying to force ambiguity in the question. I disagree. –  Feb 17 '23 at 08:57
  • Given my decade of experience reading questions on this site, that kind of ambiguity that you claim isn't there is there all the time. It sneaks in for instance when the asker makes minor changes to the phrasing of a problem that they think are inconsequential, but really aren't. Given that this is likely an exercise where they are asked to prove that $b\geq c$, and that I think it more likely that the OP made a mistake than the original problem author, then yes, I think it is apt to ask for clarification before claiming the entire problem statement is wrong. – Arthur Feb 17 '23 at 09:03
  • @Arthur: please harass someone else. –  Feb 17 '23 at 09:15