- #1

phoenixthoth

- 1,605

- 2

I'm dimly aware of what to do or where to begin.

To keep the notation straight, A[0] is the set of axioms in ZFC. A[1] is the set of axioms in ZFA, antifoundation. I know that A[0] consistent implies A[1] consistent. Finally, A[2] is A[1] plus one more axiom.

From what little I understand, what I need to do is show that if (M,I) is a model for A[1] then there is a model (M',I') of A[2].

What is the proper way to show M' exists?

FYI, I'm working with wffs in a ternary logic. Russell-type paradoxes become Q <--> ~Q which is not always false: it can be neither true nor false. The one axiom I'm adding to A[1] is a universal set axiom E(x)A(Y)y∈x. For reference, denote this by U.

BTW: We would say (N,I)|=f, read "N satisfies f" if f is true when interpreted in N with I.

So I think that M'=M∪{U'} where U' is the symbol whose interpretation is U. Then, I'd like to show a map q exists from M to M' that preserves interpretations that is 1-1. So a,b in M (i.e., a, b are sets in ZFA) and a∈b iff q(a)∈q(b) (in A[2]). And since ∈ is the only nonconstant logical symbol, that would be all we need (?). I don't know if I'm on the right track at all. The thing is I am not seeing where I need ternary logic...

Thanks in advance for any feedback.