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 $0=1$).

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.

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.