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We have all seen statements about equivalent conditions, such as If any one of the following three conditions hold, then all three conditions hold.

Are there any examples of three conditions which all hold, provided at least two of them hold? So to be slightly more concrete, given conditions (a), (b), and (c), we know that if (a) & (b) are true, then (c) is true, but if either (a) or (b) is false, then (c) need not be true. This formulation would hold up to any permutation of (a), (b), and (c).

I've given this some thought, but I really have no clue how to approach this question with rigor, although if there were a way, I feel it would be quite simple. I thought perhaps someone would know a way to demonstrate the existence or impossibility of this situation or have an explicit example.

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If any two of $(A\wedge B)$, $(A\wedge C)$, and $(B\wedge C)$ are true, then all three must be true; but none of these conditions by itself implies any other. – mjqxxxx Oct 24 '11 at 19:51
@mjqxxxx You ought to put that as an answer. I won't steal it from you or mention a substitution instance of it. I'd just add classical propositional logic as the context. – Doug Spoonwood Oct 24 '11 at 23:21
@mjqxxx, yes that is quite true, and I'm glad you pointed that out, but as an answer it is also quite unsatisfying :-) – PrimeRibeyeDeal Oct 25 '11 at 22:29
up vote 3 down vote accepted

Let $V$ be an $n$-dimensional vector space. Let $S$ be a subset of $V$.

(a) $S$ is a linearly independent set.

(b) The span of $S$ is $V$.

(c) The number of elements of $S$ is $n$.

Any two imply the third. No one implies either of the other two.

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Let $ABC$ be a triangle.

  1. $ABC$ is an isosceles triangle.
  2. There exists a right angle in $ABC$.
  3. There are two angles of $45^\circ$ degrees.

Every two imply the third, but one alone is not enough to imply any of the others.

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I love this simple one. Let $A$ be a set, $\sim$ be an equivalence relation on $A$, let $x_0\in A\land x_1\in A\land x_2\in A$.

  • $x_0\sim x_1$.
  • $x_1\sim x_2$.
  • $x_2\sim x_0$.
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I think your $x_3$ should be $x_0$. – Quinn Culver Nov 10 '11 at 14:34
@Quinn Culver: You are right. – beroal Nov 11 '11 at 15:11

Let $X$ be a topological space.

  1. $X$ is finite.
  2. $X$ is discrete.
  3. $X$ is compact Hausdorff.

No one implies another, but any two imply the third.

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A generalization inspired by mjqxxxx's comment and Shamir's Secret Sharing:

Let $f(x), g(x)\in K[x]$ be polynomials of degree $d$ and suppose $x_1, \dots x_m$ are distinct elements of the field $K$ for some $m> d$. Then any $d+1$ of the following statements together will imply any (all) of the others. $$1. f(x_1) = g(x_1)$$ $$1. f(x_2) = g(x_2)$$ $$\dots$$ $$1. f(x_m) = g(x_m)$$

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