I'm studying Set Theory on my own and I have a question about a proof.
The book I am reading wants to prove $ZF \vdash (AC)^L$ in order to prove the relative consistency of AC from ZF. I have no problems with the proof of the following theorem:


1) $ZF \vdash (V = L)^L.$


Here comes the problem: my book is a bit concise and I am not sure if the following proof of mine is correct. It would be very kind if you could check it:


We have a formula $\Phi$ which is $\exists \xi \psi$ where $\psi$ is $\Delta^{ZF}\_1$ (I do not know if you are familiar with this $\psi$ but it should be the canonical well-ordering of $L$), so from 1 we have
$$ZF \vdash \forall x \in L (\Phi \text{ Well orders } x)^L$$
which is equivalent to 
$$ZF \vdash \forall x \in L (\Phi^L \text{ Well orders } x)$$
(because the formula $\Phi^L \text{ Well orders } x$ is $\Delta\_1^{ZF}$)


which is equivalent to
$$ZF \vdash \forall x \in L (\Phi \text{ Well orders } x)$$
(because $\Phi$ is $\exists \xi \psi$ and all the ordinals are in $L$)


which is 
$$ZF \vdash WOT^L$$
(where WOT is the Well Ordering Theorem)


But also $ZF \vdash WOT \leftrightarrow AC$ 
so $ZF \vdash AC^L$.


QED?


Note: The last passage works because we have $ZF \vdash WOT^L \leftrightarrow AC^L$ because $WOT \leftrightarrow AC$ is $\Delta\_1^{ZF}$.



A proof of $ZF \vdash AC^L$
If you write out in the obvious way the equivalence $WOT\iff AC$, it won't be $\Delta\_1$; the ony reason it's $\Delta\_1^{ZF}$ is that it's provable in ZF (hence provably equivalent to $0=0$).


My general impression of your argument is that you're making things unnecessarily complicated. Once you have that ZF proves the relativizations to $L$ of the axioms of ZF and of $V=L$, then it follows by general logic (not anything specific to set theory) that "ZF + ($V=L$)" is consistent if ZF is. Since AC is provable in the theory "ZF + ($V=L$)", it follows (still by general logic) that "ZF + AC" is consistent if ZF is.


