yossell said:
If you could list Kunen's axioms, that might be helpful.
Definition II.10.1 A logical axiom of [itex]\mathcal L[/itex] is any sentence of [itex]\mathcal L[/itex] which is a universal closure (see Definition II.5.6) of a formula of one of the types listed below. Here, x,y,z, possibly with subscripts, denote arbitrary variables.
1. propositional tautologies.
2. [itex]\varphi\rightarrow\forall x\varphi[/itex], where x is not free in [itex]\varphi[/itex].
3. [itex]\forall x(\varphi\rightarrow\psi)\rightarrow(\forall x\varphi\rightarrow\forall x\psi)[/itex].
4. [itex]\forall x\varphi\rightarrow\varphi(x\rightsquigarrow\tau)[/itex], where [itex]\tau[/itex] is any term which is free for x in [itex]\varphi[/itex].
5. [itex]\varphi(x\rightsquigarrow\tau)\rightarrow\exists x\varphi[/itex], where [itex]\tau[/itex] is any term which is free for x in [itex]\varphi[/itex].
6. [itex]\forall x\lnot\varphi\leftrightarrow\lnot\exists x\varphi[/itex].
7. [itex]x=x[/itex].
8. [itex]x=y\leftrightarrow y=x[/itex]
9. [itex]x=y\land y=z\rightarrow x=z[/itex].
10. [itex]x_1=y_1\land\dots\land x_n=y_n\rightarrow (f(x_1\dots x_n)=f(y_1\dots y_n))[/itex], whenever n>0 and f is an n-place function symbol of [itex]\mathcal L[/itex].
11. [itex]x_1=y_1\land\dots\land x_n=y_n\rightarrow (p(x_1\dots x_n)\leftrightarrow p(y_1\dots y_n))[/itex], whenever n>0 and p is an n-place predicate symbol of [itex]\mathcal L[/itex].
(My comment: I don't know what [itex]\rightsquigarrow[/itex] means.)
Definition II.10.3 If [itex]\Sigma[/itex] is a set of sentences of [itex]\mathcal L[/itex], then a formal proof from [itex]\Sigma[/itex] is a finite non-empty sequence of sentences of [itex]\mathcal L[/itex], [itex]\varphi_0,\dots,\varphi_n[/itex], such that for each i, either [itex]\varphi_i\in\Sigma[/itex] or [itex]\varphi_i[/itex] is a logical axiom or for some j,k<i, [itex]\varphi_i[/itex] follows from [itex]\varphi_i,\varphi_j[/itex] by Modus Ponens (that is, [itex]\varphi_k[/itex] is [itex](\varphi_j\rightarrow\varphi_i)[/itex]). This sequence is a formal proof
of its last sentence, [itex]\varphi_n[/itex].
yossell said:
If it's formal proofs you want, the meaning of = doesn't matter - you can do the proof without knowing the meaning =. But that holds for ALL the symbols - so identity isn't special.
That's what I've been thinking this whole time, but it seemed to me when I started this thread that = was treated differently, because the extensionality axiom was written as
[tex]\forall x\forall y(\forall z(z\in x\leftrightarrow z\in y)\rightarrow x=y)[/tex]
and the arrow going in the other direction was left out with the motivation that it follows from the fact that = has the
meaning of equality. If we really need to use the meaning here, then =
is treated differently, but we can just use the logical axioms instead. Since the logical axioms list the properties of all the symbols (= in 7-11 and the others in 1), = isn't really treated differently.