Chain of implications shows equivalence of several conditions
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Pick two conditions, say, $k$ and $l$ where $k<l$. Then $k\implies k+1\implies...\implies l1\implies l$, so $k\implies l$. Further, $l\implies l+1\implies ...\implies n1 \implies n\implies 1\implies...\implies k$, so $l \implies k$. Thus, $k\iff l$ for any two conditions $k$ and $l$.
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wwww
Updated on August 15, 2022Comments

wwww 2 months
In mathematical articles, theorems frequently have the following form:
The following (conditions) are equivalent:
(first condition)
(second condition)
(third condition)
...
where several (usually more than two) conditions are listed.
Can you convince me that proving that showing the chain of implications 1 → 2 → 3 → ... → n → 1 (where n is the last condition) suffices to prove the equivalence of the n conditions?

Git Gud over 6 yearsHow convincing a proof do you need?

Git Gud over 6 yearsTake $n+1$ statements in the given condition and assume it works for $n$ statements. Use the induction hypothesis on the first $n$ statements. You already know that the $n^{\text{th}}$ statement implies the $(n+1)^{\text{th}}$ one, so you're only missing the other direction. You also know that the last statement implies the first. Now, due to the induction hypothesis the first implies the nexttolast one, completing the induction step.

Git Gud over 6 yearsI really don't think it's possible that someone who would ask this question would find this convincing.