TITLE: Axiom of choice and the equality between second-order constructible universe and HOD
QUESTION [6 upvotes]: I try to prove $L_{SO}=\mathrm{HOD}$, where $L_{SO}$ is second-order constructible universe which has similar definition with $L$ but it uses second-order definability rather than the first-order definability, and I found the answer in MO. Also, I found the referred article in the answer which is written by Myhill and Scott. 
The proof of $L_{SO}=\mathrm{HOD}$ in the answer mentioned previoisly and the article uses axiom of choice. (Exactly, it uses trichotomy for cardinals and they use it to prove $\mathrm{HOD}\subset L_{SO}$). My question is: using the axiom of choice is essential to prove $\mathrm{HOD}\subset L_{SO}$? 
Thanks for any information or clarification.

REPLY [8 votes]: The equality $L_{SO}=HOD$ can not be proved just in $ZF$. This is proved in the paper ``The consistency of the theory $ZF+L^1\neq HOD$'' by Szczepaniak.
Here $L^1$ refers to what you named $L_{SO}$. The idea of the proof is as follows:
$(1)$ If two models of $ZF$ have the same sets of ordinals, then they have the same classes $L^1,$
$(2)$ There are models $N_1 \subset N_2$ with the same sets of ordinals, such that there is a real $a\in N_1$ such that $a\notin HOD^{N_1}$ but $a\in HOD^{N_2}.$
Now the result follows from $(1)$ and $(2)$. The paper can be find here