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In D. Scott: More on the axiom of extensionality, in Essays on the foundations of mathematics, dedicated to A. A. Fraenkel on his seventieth anniversary, edited by Y. Bar-Hillel, E. I. J. Poznanski, M. O. Rabin, and A. Robinson for The Hebrew University of Jerusalem, Magnes Press, Jerusalem 1961, pp. 115--131, Dana Scott introduces a system $ZF^s$: Let $R^s$ be replacement with coextensionality in place of identity. $ZF^s$ is $ZF$ minus extensionality, with $R^s$ instead of ordinary replacement. Scott shows that $ZF^s$ interprets ZF.

May we show that $ZF^s$ proves that the transitive closure of any set is as well a set, without invoking the interpretation?

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    $\begingroup$ what do you exactly mean by the transitive closure? As long as Extensionality is lost, there is no quarantee that there is a unique transitive closure for a set, for each set there can be many transitive closures and all of them are of course co-extensional. If you meant to prove existence of a transitive closure for each set, then this is easy, its moreorless the same traditional proof with some modification to follow co-extensionality in replacing from $\omega$ to the $n$-unions instead of identity as usual. $\endgroup$
    – Zuhair Al-Johar
    Commented Sep 15, 2021 at 18:04
  • $\begingroup$ @ZuhairAl-Johar Yes, I was looking for at least one transitive closure for each set. If you may help me state an easy solution, without using the recursion theorem, I would be grateful. $\endgroup$
    – Frode Alfson Bjørdal
    Commented Sep 17, 2021 at 4:29
  • $\begingroup$ @ZuhairAl-Johar I hope the following interests you: mathoverflow.net/questions/404362/… $\endgroup$
    – Frode Alfson Bjørdal
    Commented Sep 20, 2021 at 22:15
  • $\begingroup$ @ZuhairAl-Johar The construction will work without extensionality as well, if the goal is to achieve a transitive closure. $\endgroup$
    – Frode Alfson Bjørdal
    Commented Sep 20, 2021 at 22:16

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