I was reading up on Pragmatism and the pragmatic maxim, which states: "Consider the practical effects of the objects of your conception. Then, your conception of those effects is the whole of your conception of the object".

I thought that it might be interesting to generalize this to mathematics, with some alterations.

So, here is my question:

"Let B be the set of all implications of a proposition $a$. Then there exists a function such that $f(a)=B-a.$ Is this function injective?"


1 Answer 1


$\mathsf{Prop}$ with $\land$, $\lor$, $\lnot$ forms a boolean lattice, and the ordering is given by $x \leq y \iff$ $x \to y$.

Then asking if a propostion $a$ is determined by $\{ b \neq a ~|~ a \to b \}$ is asking if $a$ is determined by $\{ b > a \}$.

This is not true, though. For instance consider the lattice with four elements


Then $a$ and $\lnot a$ are both sent to $\{ T \}$ by your function, so it is not injective in general.

If we look at $\{b \geq a \}$ instead, which is your function but without removing $a$ from the set, we recover injectivity (for instance, by the Yoneda Lemma).

I hope this helps ^_^

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    $\begingroup$ +1. At the same time, there are some theories relative to which the result does hold - e.g. first-order Peano arithmetic or ZFC. $\endgroup$ Dec 31, 2020 at 14:27
  • $\begingroup$ @NoahSchweber Can you elaborate on the proof of that? $\endgroup$ Jan 2, 2021 at 17:20
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    $\begingroup$ @JohnClever See e.g. the first part of this old answer of mine (that proves a particular "downward-density" result, but the same argument gives the more general result in my comment). $\endgroup$ Jan 2, 2021 at 17:23
  • $\begingroup$ Thank you, this looks helpful, and gives me another reason to get into algebraic logic. $\endgroup$ Jan 2, 2021 at 17:32
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    $\begingroup$ @JohnClever - If this answered your question, you should mark it as such so that other users know how to spend their time ^_^ $\endgroup$ Jan 5, 2021 at 7:37

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