Indicator function with respect to set complement

Let $X$ be a D11: Set such that
 (i) $E \subseteq X$ is a D78: Subset of $X$ (ii) $X \setminus E$ is the D79: Set complement of $E$ within $X$ (iii) $I_{X \setminus E} : X \to \{ 0, 1 \}$ is an D41: Indicator function on $X$ with respect to $X \setminus E$
Then $$I_{X \setminus E} = 1 - I_E$$
Proofs

Let $X$ be a D11: Set such that
 (i) $E \subseteq X$ is a D78: Subset of $X$ (ii) $X \setminus E$ is the D79: Set complement of $E$ within $X$ (iii) $I_{X \setminus E} : X \to \{ 0, 1 \}$ is an D41: Indicator function on $X$ with respect to $X \setminus E$
We understand both functions $I_E$ and $I_{X \setminus E}$ as functions from $X$ to $\{ 0, 1 \}$. Since the domain and codomain sets are the same, it suffices to show that they attain the same values on $X$.

If $X$ is empty, then the equality $I_{X \setminus E}(x) = 1 - I_E(x)$ holds vacuously for all $x \in X$, so assume that $X$ is nonempty and let $x \in X$. If $x \in E$, then $$1 - I_E(x) = 1 - 1 = 0 = I_{X \setminus E}(x)$$ If instead $x \in X \setminus E$, then $$1 - I_E(x) = 1 - 0 = 1 = I_{X \setminus E}$$ Result R977: Ambient set is union of subset and complement of subset shows that $X = E \cup (X \setminus E)$. That is, these two cases exhaust all possible scenarios. Since $x \in X$ was arbitrary, the claim is established. $\square$