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](https://mathoverflow.net/a/156949/48041). 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. (To be precise, 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.



Axiom of choice and the equality between second-order constructible universe and HOD
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](http://link.springer.com/chapter/10.1007%2FBFb0067659)


