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If we add a primitive binary relation $\sim$ to denote "bisimilarity" relation. Remove the axiom of Extensionality from axioms of $\sf ZFC$, and add:

Bisimilarity: $\forall x \, (x \sim x) \\ \begin{align} \forall x \forall y \, (x \sim y \iff &\forall u \in x \exists w \in y \, (w \sim u) \land \\&\forall w \in y \, \exists u \in x \, (u \sim w)) \end{align} $

Define: $\operatorname {Set}(x) \iff \forall a,b \in x: a \neq b \to a \not \sim b $

We alter the axiom of Power sets to:

Power': $\forall x \exists y: \operatorname {Set}(y) \land \forall z \subseteq x \exists {\sf z} \in y\, ( {\sf z }\sim z)$

Infinity': $\exists x: \operatorname {Set} (x) \land 0 \in x \land \\\forall y \in x \exists z \in x: \forall m (m \in z \leftrightarrow m=y)$

Also replace Replacement by Collection, and keep other axioms of $\sf ZFC$.

Add the following axioms:

Reduction: $\forall a \exists \alpha: \operatorname {Set}(\alpha): a \sim \alpha$

De-Extensionality: $\forall x \not \exists y: \forall {\sf x} (x \equiv {\sf x} \to {\sf x} \in y)$

Where $\equiv$ is coextensionality relation, i.e. having the same members.

Is $\sf ZFC$ interpretable in this system?

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  • $\begingroup$ I'm not a Set Theorist, so don't quote me on this, but I think Dana Scott's 1966 paper on Extensionality showed that ZF without it is provably consistent in Zermelo set theory alone. That is, no! $\endgroup$ Commented Feb 6, 2023 at 19:45
  • $\begingroup$ Just for curiosity, what motivates your question? $\endgroup$
    – Hanul Jeon
    Commented Feb 7, 2023 at 4:12
  • $\begingroup$ @Hanul Jeon, In reality an approach to NF through bisimulation as an alternative to Extensionality, but I need to understand it first in ZF context, since models of NFU can be retrieved through automorphisms from models of ZF, so I need to make sure that it works in ZF. $\endgroup$ Commented Feb 7, 2023 at 8:43
  • $\begingroup$ @PaulTaylor Yes. That is the case when the usual Replacement axiom schema is used. But, when Collection is used instead, then it can interpret full ZFC. But, the situation here is different. $\endgroup$ Commented Feb 7, 2023 at 11:41
  • $\begingroup$ If understanding bisimulation is the only purpose, I would rather suggest Aczel's Non-well-founded set or Harvey Friedman's interpretation of $\mathsf{ZF}$ to $\mathsf{IZF}$. $\endgroup$
    – Hanul Jeon
    Commented Feb 7, 2023 at 17:54

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