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Sep 13 at 15:34 comment added Joel David Hamkins It seems delicate to interpret ZFC in your theory, since the interpreted model will have Dedekind infinite sets, but to prevent those sets from giving rise to actual Dedekind infinite sets will mean that the interpretation must be using an equivalence relation (i.e. interpreting = nontrivially).
Sep 13 at 15:32 comment added Joel David Hamkins Interpreting ZFC is equivalent to interpreting ZFC+V=L, since L is definable in ZFC and so every interpreted model of ZFC also interprets a model of ZFC+V=L. Your theory doesn't seem to have its own L (unless this is really what you are asking?), so I think the L angle of your second question is irrelevant.
Sep 13 at 15:00 history edited Zuhair Al-Johar CC BY-SA 4.0
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Sep 13 at 15:00 comment added Zuhair Al-Johar @JoelDavidHamkins, Ah! OK, I'll correct it.
Sep 13 at 14:10 comment added Joel David Hamkins But then you wouldn't be constructing $L$, but interpreting a model.
Sep 13 at 12:56 history edited Zuhair Al-Johar CC BY-SA 4.0
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Sep 13 at 12:49 comment added Zuhair Al-Johar @JoelDavidHamkins, I thought $\omega$ could be interpreted, even though not provable in the system.
Sep 13 at 12:32 comment added Joel David Hamkins What do you mean by constructing $L$ if you don't have $\omega$?
Sep 13 at 12:26 comment added Zuhair Al-Johar @EmilJeřábek, OK, I've changed the title to better suit the question.
Sep 13 at 12:26 history edited Zuhair Al-Johar CC BY-SA 4.0
edited title
Sep 13 at 11:50 comment added Emil Jeřábek The answer to the question in the title, which is quite different from the question in the question, is yes: e.g., take $\mathrm{ZF_{fin}+Con_{ZFC}}$.
Sep 13 at 11:26 history asked Zuhair Al-Johar CC BY-SA 4.0