Symmetry of the Batalin-Vilkovisky (BV) antibracket operation Batalin and Vilkovisky define $^1$ an operation they call antibracket which is
$$(F,H)
=
\Big(\frac{\partial_r F}{\partial \Phi^A}\Big)
\Big(\frac{\partial_l H}{\partial \Phi^* _A} \Big)
-
\Big(\frac{\partial_r F}{\partial \Phi^* _A}\Big)
\Big(\frac{\partial_l H}{\partial \Phi^A} \Big) \tag{0}
$$
where index $r$ and $l$ indicate respectively right and left derivative with respect to fields $\Phi$ and antifield $\Phi^*$.
This operation has some similar properties with super-Lie commutator such as Grassmann Parity
$$
 \epsilon[(F,H)] = \epsilon(F) + \epsilon(H) + 1\tag{1}
$$
and symmetry
$$
 (F,H) = -(-1)^{(\epsilon_F + 1)(\epsilon_H+1)}(H,F)\tag{2}
.$$
My question is about these properties. I am working on the way to prove them. I tried to use that $$FH = (-1)^{\epsilon_F\epsilon_H}HF\tag{3}$$ in second property but I didn't get to the right answer. 
--
$^1$ I. A. Batalin and G. A. Vilkovisky, Phys. Lett. B102 (1981) 27.
 A: *

*Fun fact: Eqs. (1) & (2) can be intuitively understood via the Koszul sign rule/convention as that the comma in the antibracket (0) is Grassmann-odd/Fermionic $|,|=1$. E.g. the parity for permuting $F   \leftrightarrow ","\leftrightarrow H$ in eq. (2) is then after some Boolean algebra$^1$
$$|F||H| + |F||,| +|H||,| ~=~|F||H| + |F| +|H| ~=~ (|F|+1)(|H|+1)+1. \tag{A}$$
Of course eq. (A) is only a pseudo-explanation. Let's now turn to an actual proof. 

*Sketched proof of eq. (2): Use the following rules:
$$ |\Phi^{\ast}_A|~=~|\Phi^A|+1~:=~|A|+1, \tag{B}$$
and
$$ \frac{\partial^LF}{\partial z}~=~(-1)^{(|F|+1)|z|}\frac{\partial^RF}{\partial z}. \tag{C}$$
So e.g. the parity for permuting the first term
$$ \frac{\partial^RF}{\partial \Phi^A}\frac{\partial^LH}{\partial \Phi^{\ast}_A}  \leftrightarrow  \frac{\partial^RH}{\partial \Phi^{\ast}_A}\frac{\partial^LF}{\partial \Phi^A}\tag{D}$$ 
in the antibracket (0) is
$$ \underbrace{\overbrace{(|F|+|A|)(|H|+|A|+1)}^{\text{rule } (3)}
}_{=~(|F|+|H|)|A|+|F|(|H|+1)} 
+\underbrace{\overbrace{(|F|+1)|A|}^{\text{rule } (C)} 
+\overbrace{(|H|+1)(|A|+1)}^{\text{rule } (C)}
}_{=~(|F|+|H|)|A|+(|H|+1)}
~=~(|F|+1)(|H|+1).\tag{E}$$
Note that the rhs. (E) does not depend on $|A|$ (as it shouldn't!), and that the rhs. (E) is opposite the sought-for result (A). This explains the relative sign between the 2 terms in the antibracket (0). 
$\Box$
--
$^1$ In this answer we use 
$$|\cdot|~\equiv~ \epsilon(\cdot)~\in~ \mathbb{Z}/2\mathbb{Z}~\cong~\{0,1\} \tag{F}$$ 
to denote Grassmann-parity.
