# Are propositions uniquely determined by their non-equivalent implications?

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?"

$$\mathsf{Prop}$$ 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 ^_^

• +1. At the same time, there are some theories relative to which the result does hold - e.g. first-order Peano arithmetic or ZFC. Dec 31, 2020 at 14:27
• @NoahSchweber Can you elaborate on the proof of that? Jan 2, 2021 at 17:20
• @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). Jan 2, 2021 at 17:23
• Thank you, this looks helpful, and gives me another reason to get into algebraic logic. Jan 2, 2021 at 17:32
• @JohnClever - If this answered your question, you should mark it as such so that other users know how to spend their time ^_^ Jan 5, 2021 at 7:37