The classical Davis inequality $\mathbb{E} Mf\simeq \mathbb{E} Sf$, where $(Sf)^2=\sum_{k}\left|f_{k}-f_{k-1}\right|^2$ is the square function and $Mf= \sup_n \left|f_n\right|$ is the maximal function, is true with a universal constant for any martingale $f$ on any filtration. A natural analog in the setting of (F4) doubly indexed filtrations, i.e. $\left(\mathcal{F}_{i,j}\right)_{i,j}$ such that the operators $\mathbb{E}\left(\cdot\mid \mathcal{F}_{i,\infty}\right)$ and $\mathbb{E}\left(\cdot\mid \mathcal{F}_{\infty,j}\right)$ commute and their product is $\mathbb{E}\left(\cdot\mid \mathcal{F}_{i,j}\right)$, is the conjecture \[\mathbb{E}\sup_{n,m} \left|f_{n,m}\right|\simeq\mathbb{E}\left(\sum_{i,j}\left|Δf_{i,j}\right|^2\right)^\frac{1}{2},\] where $Δf_{i,j}=f_{i,j}-f_{i-1,j}-f_{i,j-1}+f_{i-1,j-1}$. It was known to be true only with some highly restrictive additional assumptions, e.g. regularity of the filtration ($g_{n,m}\gtrsim g_{n+1,m},g_{n,m+1}$ for any positive martingale $g$) or $f$ being a strong martingale ($\mathbb{E}\left(Δf_{i,j}\mid \mathcal{F}_{i-1,j}\vee \mathcal{F}_{i,j-1}\right)=0$). We prove the inequality $\lesssim $ assuming just the (F4) condition.