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I'm currently working on an analytical philosophy problem, and I think formal logic tools could help me. First of all, I introduce some definitions. Let us give ourselves a language $\mathcal{L}$ and a theory $\mathcal{T}$. Unless otherwise stated, we place ourselves within this theory, and do not deviate from it.

Let us take a property (or first order predicate) $p$ such that $\neg p$ is not a tautology. I call "$p$-object" any object $x$ such as $p(x)$. My intuition is as follows: I would like to construct a p-object (which I call $\pi(p)$) which is "minimal", i.e. which has the "fewest possible properties". More formally, I want that if $\pi(p)$ has the property $q$, then $q=p$ or $q$ is a consequence of $p$ (which is inevitable by modus ponens) or $q $ is a "universal" property that all objects have (i.e. a tautology). These three conditions are united under the term $\forall x, p(x) \rightarrow q(x)$. In other words, I would like to construct an object $x$ such that: $$(\varphi): \text{ for any property q, } q(x) \rightarrow (\forall y, p(y) \rightarrow q(y))$$

Now we already have a first problem which I will expose. Suppose that for any property $p$ such that $\neg p$ is not a tautology (otherwise then $\not\exists x : p(x)$ and $\pi(p)$ cannot exist), there exists a coherent object satisfying the axiom $\varphi$. Let us take a property $p$ such that $\neg p$ is not a tautology, and let $\pi$ be a $p$-object responding to $\varphi$. Let us define $$q_p(y) := "\text{for any property r, } r(y) \rightarrow (\forall z, p(z) \rightarrow r(z))"$$

So by definition we have $q_p(\pi)$. So the object $\pi$ has the property $q_p$, and yet we will show that there exists a $p$-object which is not a $q_p$-object; and therefore that $\pi$ does not in fact respect the axiom $\varphi$. Indeed, $\exists z : p(z) \wedge \neg q_p(z)$.

Proof.

Let $s$ be a property such that $\neg (s \wedge p)$ is not a tautology (so that $\pi(s \wedge p)$ can exist) and such that $p \rightarrow s$ is not either. By instanciation this means that $\exists x : (p(x) \wedge \neg s(x))$. We then have $$\begin{align} &\neg q_p(\pi(s \wedge p)) \\ &\leftrightarrow \text{There is a property r such that } r(\pi(s \wedge p)) \wedge (\exists z : p(z) \wedge \neg r(z)) \\ &\leftarrow s(\pi(p \wedge s)) \wedge \exists x : (p(x) \wedge \neg s(x)) \\ &\leftarrow (p \wedge s)(\pi(p \wedge s)) \wedge \exists x : (p(x) \wedge \neg s(x)) \\ &\leftrightarrow \textbf{1} \wedge \textbf{1} \\ &\leftrightarrow \textbf{1} \end{align}$$ So we have $\textbf{1} \rightarrow \neg q_p(\pi(s \wedge p))$ therefore $\neg q_p(\pi(s \wedge p))$ but by definition we have $(p \wedge s)(\pi(p \wedge s))$ so we have $p(\pi(p \wedge s))$.

QED.

In other words, $\pi$ has the property $q_p$, but the latter is not universal, is not $p$, and is not a consequence of $p$. In reality, the problem is more simply the following: the object defined by "the object having as sole property only the consequences of $p$"... does not have as sole property the consequences of $p$ ! In fact its definition constitutes a property which does not follow from $p$, but which this object nevertheless has. We could resolve the problem as follows: instead of defining "the object whose only properties are the consequences of $p$"; we define instead "the minimal $p$-object is the object having as sole properties the consequences of $p$, and the consequences of its minimality".

I would like to give some ideas and some comments. To begin with, we can notice that in reality all objects have the same number of properties since not having a property is a property. Indeed, we have $\neg (p(x)) \leftrightarrow (\neg p)(x)$. This therefore poses a problem when it comes to choosing a $p$-object having, intuitively, "the fewest possible properties". However, the solution is self-evident: it is necessary to determine “affirmative” properties and “negative” properties. For example, $"x=0"$ is an affirmative property and $"x \neq 1"$ is a negative property. That said, is the property $A \wedge \neg B$ negative or affirmative? It is impossible to conclude, and we must then choose "case by case" which are the affirmative properties, and deduce from this that their negation are exactly the negative properties.

We start by noting that the properties are elements of $\mathcal{F} \left( \textbf{Obj}_\mathcal{T}, \mathcal{L}^{(\mathbb{N})} \right )$ since they are applications taking as input an object of the theory (i.e. a symbol-object of the language) and as output a closed formula (i.e. a finite sequence of symbols of the alphabet). We can then define $\textbf{Prop}^1$ the set of first-order predicates in our theory. We introduce an equivalence relation on this set: $$\forall p,q: \textbf{Prop}^1, p \asymp q \text{ iff. } (\forall x, p(x) \leftrightarrow q(x)) \vee (\forall x, p(x) \leftrightarrow \neg q(x))$$. By axiom of choice, we choose a system of representatives of $\asymp$ which we call $\textbf{Aff}$. We are seeking to define, for a first order predicate $p$ (such that $\neg p$ is not a tautology), an object $\pi_\textbf{Aff}(p)$. We would then like to show that such an object is unique (and that giving it a single name is justified), and why not finally that it does not in fact depend on $\textbf{Aff}$.

A possible way of doing it would be as follows (I specify that I have not studied formal logic, nor set (or class) theory, I do not know enough about it to avoid possible problems of "false sets" like $\{x \mid x \not\in x\}$. Forgive me if I make such mistakes):

We define $$\forall x, \text{CoDom}^+(x) := \{p: \textbf{Aff} \mid p(x) \}$$ the set of affirmative properties of $x$. We then define, for any collectivizing property $p: \textbf{Prop}^1$ (i.e. such that $\{x \mid p(x)\}$ is a set), its "domain" as: $$\text{Dom}(p) := \{ x \mid p(x) \}$$

And we define a partial order relation on the domain of $p$ as follows: we will write, $\forall x,y \in \text{Dom}(p)$, $$x \leq y \text{ iff . } \text{CoDom}^+(x) \subseteq \text{CoDom}^+(y)$$ This makes the domain of $p$ a partially ordered set of $p$-objects. From there, we could maybe use Zorn's lemma to find the existence of smaller elements in $\text{Dom}(p)$. Note that this strategy only defines (and it is perhaps better that way; although I wonder if it is necessary) $\pi_\textbf{Aff}(p)$ for collectivizing $p$. Using Zorn's lemma would actually require, for a totally ordered set $E \subseteq \text{Dom}(p)$, to find a $p$-object $x$ such that $\forall y \in E, x \leq y$. I personally don't really see how to do this without already assuming the existence of a minimal $p$-object... Actually, it's easy to determine that $$\forall x \in E, \bigcap_{y \in E} \text{CoDom}^+(y) \subseteq \text{CoDom}^+(x)$$ But then we would need to find a $p$-object $x$ such that $$\text{CoDom}^+(x) = \bigcap_{y \in E} \text{CoDom}^+(y)$$ Which seems difficult to me, because $\text{CoDom}^+$ is not (I think) surjective (I however think that it is injective, by ontological definition of Leibniz's concept of identity : two objects are equal if and only if they are indistinguishable, in the sense that they have exactly the same properties). Here $x$ would be, in a sort, a "intersection of objects", if it has any sense. Another idea would maybe to use the categorical concept of "free object" but my knowledge in category theory is too slim to do that !

There you go, I think I've pretty much said it all. So my questions are, to conclude:

1 - Are the $\pi(p)$ objects correctly defined, unique and consistent? If not, can they be defined in another way? Even if it means changing the properties a little, as long as the intuition of the "minimal" $p$-object remains...

2 - Do they depend on $\textbf{Aff}$, or even on something else external to $p$?

3 - If they exist, for what propositions $p$ do they exist?

I thank you all for reading, and thank you in advance for your answers, I can't wait to read them! Have a good day !

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  • $\begingroup$ Alas, your question is a poor fit for the site in its current form: as the close votes say, questions here need a much tighter focus. Generally, "I made a large number of observations and new definitions and now need some input" won't get much engagement. But if you're looking for inspiration / references to similar ideas or concepts that appear in the literature, you should check out generic points and free objects. $\endgroup$
    – Z. A. K.
    Feb 28 at 14:02

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