Is a mathematical statement uniquely determined by all of its necessary conditions?

597 Views Asked by At

The question I have is basically the one in the title: if I fix a statement $P$, is the set of all conditions $\{X_i\}$ necessary for $P$ to be true enough information for me to recover $P$?

An attempt to make the question more precise is as follows. Fix a formula $P$ in some language $\mathcal{L}$. Given a set of formulas $\{X_i\}_{i\in\mathcal{I}}$ such that the set $\{P\rightarrow X_i\}_{i\in\mathcal{I}}$ is maximally consistent, can I determine $P$ uniquely up to logical equivalence?

As an example where this is true, consider the first order language $\mathcal{L}_\text{NT}$ of number theory, and the following two statements

$$X_1 \equiv \text{"$x$ is even"} \quad X_2 \equiv \text{"$x$ is prime"}.$$

from these necessary conditions, I can determine that, for any set of conditions $\{X_i\}$ containing $X_1$ and $X_2$, if all $\{P\rightarrow X_i\}$ are simultaneously satisfied, then, up to logical equivalence, we have $P \equiv \text{"$x = 2$"}$.

2

There are 2 best solutions below

8
On BEST ANSWER

Up to provable equivalence, the answer is "yes"!

Every first order theory can be associated to a syntactic category where formulas $\varphi(x_1, \ldots x_n)$ correspond to subobjects of an object $X^n$ (here you should think of $X$ as the "underlying set" of a "free model" of your theory. Then the formula $\varphi(x_1, \ldots, x_n)$ is exactly the subobject $\{ (x_1, \ldots, x_n) \mid \varphi(x_1, \ldots, x_n) \} \subseteq X^n$. This is a little white lie, but is morally correct).

Now the (definable) subobjects of $X^n$ (up to provable equivalence) form a lattice, and $\phi \leq \psi$ happens if and only if $\phi \vdash \psi$! Now what does yoneda's lemma tell us for this lattice? It says exactly that $\psi$ is determined up to isomorphism (that is, up to provable equivalence) by elements below $\psi$.

So $\psi(x_1, \ldots, x_n)$ is determined up to provable equivalence by

$$ \{ \phi(x_1, \ldots, x_n) \mid \phi(x_1, \ldots, x_n) \vdash \psi(x_1, \ldots, x_n) \} $$

Or, if you like,

$$ \{ \phi(x_1, \ldots, x_n) \mid \phi(x_1, \ldots, x_n) \to \psi(x_1, \ldots, x_n) \} $$

since $\phi \vdash \psi$ implies $\phi \to \psi$ is the top element of the lattice (True).

For more info, you might be interested in Steve Awodey's lecture notes here.


I hope this helps ^_^

8
On

I like the category-theoretic perspective in HallaSurvivor's answer, but let me give a more "down to earth" version of the same argument, which might help some readers.

Fix an $L$-sentence $P$. Let $N = \{Q\mid Q\text{ is an $L$-sentence, and }P\vdash Q\}$, the set of all $L$-sentences which are necessary for $P$. I claim that $P$ can be recovered, up to logical equivalence, from $N$. Indeed, $P$ is characterized up to logical equivalence as the sentence in $N$ with the property that it is sufficient for every sentence in $N$. That is, by the following properties: (1) $X\in N$, and (2) for all $Q\in N$, $X\vdash Q$.

First note that $P$ satisfies (1) and (2). For (1), $P\vdash P$, so $P\in N$. For (2), for all $Q\in N$, $P\vdash Q$ by definition of $N$.

Now suppose $P'$ satisfies (1) and (2). I claim that $P'$ is logically equivalent to $P$. By (1), $P'\in N$, so $P\vdash P'$. And since $P\in N$, by (2), $P'\vdash P$. So $P$ and $P'$ are logically equivalent.


I believe this answers the question as you have stated it in your title and first paragraph. Unfortunately, the rest of what you wrote (in particular, the "attempt to make the question more precise") makes little sense to me. Here are some clarifying questions:

  • In the title question, you talk about all conditions necessary for $P$. But in the later paragraphs, you look at an arbitrary set of conditions $X_i$ which are necessary for $P$. Which one did you mean?
  • What exactly do you mean by "the set $\{P\rightarrow X_i\}_{i\in \mathcal{I}}$ is maximally consistent"? A set of conditional sentences of this form will never be a maximally consistent (i.e., complete) theory unless $P$ is logically valid. But maybe you meant maximally consistent among sets of sentences of this form? Note that (unless $P$ is valid), any set of sentences of this form is consistent, since they're all true when $P$ is false.
  • In the next paragraph, you switch out "the set $\{P\rightarrow X_i\}_{i\in \mathcal{I}}$ is maximally consistent" for "the set $\{P\rightarrow X_i\}_{i\in \mathcal{I}}$ is simultaneously satisfied". As Andrej Bauer points out in the comments, there's another possibility where $P\rightarrow X_1$ and $P\rightarrow X_2$ are simultaneously satisfied, namely when $P$ is contradictory: $\bot$, or $x\neq x$.