Denote $\bar{A}$ a complement of $A$ in a set $\Omega$ and $A \Delta B = A/B \cup B/A$ the symmetric difference of $A, B$. It is claimed that for a map $\phi := \Omega \rightarrow \lbrace 0, 1 \rbrace, \; \phi(\emptyset) = 0$:
$$ \begin{array}{|c|c|} \hline \phi(A) & \phi(\bar{A}) \\ \hline 0 & 1 \\ 1 & 0 \\ \hline \end{array}, \, A \in \Omega \implies \phi(A \Delta B) = \phi(A) + \phi(B), \text{ and } \phi(\Omega) = 1 $$
I know of a counterexample to this. Let $\mathcal{P}(\alpha)$ be a set of all subsets (a power set) of set $\alpha$. Consider $\phi: \Omega \rightarrow \lbrace 0,1 \rbrace$, $\Omega = \mathcal{P}({\lbrace a, b, c \rbrace})$:
$$ \begin{align*} \phi(a) &= 0 & \phi(b) &= 1 & \phi(c) &= 1 & \phi(\emptyset) &= 0 \\ \phi(b \cup c ) &= 1 & \phi(a \cup c) &= 0 & \phi(a \cup b) &= 0 & \phi(a \cup b \cup c) &= 1 \end{align*} $$
which satisfies the truth table, but breaks linearity ($\phi(a \Delta c) = \phi(a \cup b) = 0 \neq \phi(a) + \phi(c) = 1$).
I am interested if the following could be also implied by the LHS and also lead to a valid counter:
$$ \phi(A \Delta B) = \phi(A) + \phi(B) + \phi(A)\phi(B), \, \text{ and } \phi(\Omega) =1 $$
Choosing $\phi(A) = 0$ implies that: $$ \phi(A \Delta \bar{A}) = \phi(\Omega) = 1 = \phi(A) + \phi(\bar{A})(1 + \phi(A)) = \phi(\bar{A})$$ and hence: $\phi(A) = 0 \implies\phi(\bar{A}) = 1$ and $\phi(\bar{A}) = 0 \implies \phi(A) = 1$.
Choosing $\phi(A) = 1$ however leads to: $$ 1 = 1 + \phi(\bar{A})\cdot 0 = 1 $$ no matter what $\phi(\bar{A})$ is, so the opposite implication is not simply shown to hold. Is there any way out of this to finish the counter?
Note: If you can think of a better title for the question, please let me know.