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The following theory is a class theory that combines two principles that of reflection and resemblance, informally it says that the class $V$ of all sets resembles a set $W$ that stands as a sub-world in the sense that $W$ has the same internal structure describable in the pure language of class theory (i.e. not using $W$) that $V$ has, and in addtion $W$ would reflect inside it all existential properties of sets describable in the pure language of class theory from parameters in $W$

Two questions:

  1. Is this theory consistent?

  2. If 1. is positively answered, then is Resemblance schema independent of the other axioms?

Formally speaking the theory is formulated in mono-sorted first order predicate logic with primitive symbols of $``=, \in, W"$, standing for "identity, membership, and Sub-world", where the last is a constant symbol.

Axioms are those of identity theory +

Extensionaltiy: $\forall x (x \in a \leftrightarrow x \in b) \to a=b$

Define: $set(x) \iff \exists y (x \in y)$

Class Comprehension: if $\varphi(y)$ is a formula in which $x$ is not free, then: $[\exists x \forall y (y \in x \leftrightarrow set(y) \wedge \varphi(y))]$ is an axiom.

Let $V$ be the class of all sets.

Subworld: $W \in V$

Reflection: if $\varphi$ is a formula in $L(=,\in)$, in which $x$ occurs free, with parameters $\vec{p}$, then: $$\vec{p} \in W \to [\exists x \in V (\varphi) \to \exists x \in W (\varphi)]$$ is an axiom.

Resemblance: if $\varphi^X$ is a formula in prenex normal form, whose matrix is in $L(=,\in)$, and each variable in its prefix must appear as bounded either $\in X$ or $\subseteq X$, then: $$\varphi^V \leftrightarrow \varphi^W $$, is an axiom.

Size limitation: $\forall x (x \in V \leftrightarrow |x|<|V|)$

Foundation over all classes.

I personally tend to think that if this theory is consistent, then the model spoken about in the answer to this question, would satisfy this theory. But I'm not sure of that.

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First, observe that $V$ is transitive (By Foundation. Specifically $V=V_{Ord}$) and so $W$ is infinite. Specifically, If $|V|=n$, then $|W|=n$, and $|V|=n+1$. As W is infinite, by Global Choice there is some ordinal bijective to it (We can find some well order $R$ of $W$, and then an isomprhisim $(α,<)$ to $(W,R)$, which is by definition a bijection).

As $ω≤α$, $V$ satisfies infinity. As $2<ω$, an $\{a,b\}$ is the image of $2$, $V$ satisfies pairing. From Foundation we have that, for every $X$, $rank(X)$ is a set and also $rank(\{X,X\})>rank(X)$ and so $Ord$ is a limit and therefore $V=V_{Ord}$ satisfies union and powerset.

Therefore $V_{Ord+1}$ satisfies $KM$. However, Resemblence is really second order Reflection. As a consequence, $2^W\vDash KM$, and so your theory proves $Con(KM)$. However, consider the model $V_{\kappa+1}$, where $\kappa$ is a $1-$indescribable cardinal. We can find some $V_\lambda$ that is a second order elementary substructure of $V=V_\kappa$, and so $V_\lambda=W$ satisfies your theory. Therefore your theory is consistent relative to large cardinals.

As for part 2, we can construct a truth predicate for first order truth $T_0$ in $KM$. Therefore we can find some $W$ that satisfies the same first order sentences (Reflection) as $V$. Conversely $KM$ can be derived from your theory (As shown above), so without Resemblance the theories are in fact equivalent.

With Resemblance you can prove $Con(KM)$ and so (If your theory is $T$), by Gödel's incompleteness theorems Resemblance cannot be proved consistent from to the rest of $T$.

Edit: If you are looking for exact consistency strength, then I claim the models $M$ of $T$ are precisely those sets of the form $M=V_{\kappa+1}$, where $\kappa$ is $1-$indescribable. I discussed the backwards implication above, for the forward it is immediate that $M=V_{\kappa+1}$. We can use the reflection theorem (Not the axiom) to find arbitrarily large $\lambda$ such that $V_\lambda\prec_1 V$, by using the truth predicate $T_0$ for $W$ as a parameter. Getting the same thing to work for $S\subseteq V$ obeys a similar arguement.

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  • $\begingroup$ Ok, the theory is consistent! But can you tell me what's the exact consistency strength of this theory? What large cardinal properties it can define (up to.. which?) $\endgroup$ – Zuhair Jul 15 at 7:11
  • $\begingroup$ You seem to think lately that most Ackermann\MK versions class theories that I've presented here has consistency strength of ORD is mahlo. Do you think its the case here as well. $\endgroup$ – Zuhair Jul 15 at 9:03
  • $\begingroup$ I know. "$ORD$ is Mahlo" is everywhere. You really won't ever be able to get past "$ORD$ is Mahlo" using standard Ackermann based set theories... That being said, I think this does surpass "$ORD$ is Mahlo," give that it is not a standard Acerkmann based set theory. I will add it to my post. $\endgroup$ – Master Jul 15 at 15:36

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