Let $X$ be a set and $f$ be a function from $X$ to $\{0, 1\}$, the field with two elements. The support of $f$ is the set $f ^ {-1}$ (1), which we denote by $\DeclareMathOperator{\supp}{supp}\supp(f)$. For a collection $F$ of such functions define $\:\supp(F) = \bigcap_{f∈F} \supp(f)$
Note that when $F = ∅, \:\supp(F) = X$, and that for such functions, $f$ is the characteristic function of its support, i.e., $f = χ_{\supp(f)}$. Find the supports of $f · g$, $1 − f$, $f + g − f · g$, and $1 − f + f · g$ in terms of the supports of $f$ and $g$.
Do now know where to start. Will be helpful to get at least the first one to see how it works.
I give you a couple of examples.
Let $f \cdot g \colon X \to \{0,1\}$ be the function defined by $(f \cdot g) (x) = f(x) \cdot g(x)$ for all $x \in X$. The support of $f \cdot g$ is the intersection of the supports of $f$ and $g$. Indeed, by definition of support, \begin{align} \operatorname{supp}(f \cdot g) &= \{x \in X \mid f(x) \cdot g(x) = 1\} \\ &= \{x \in X \mid f(x) = 1 \text{ and } g(x) = 1\} \\ &= \{x \in X \mid f(x) = 1\} \cap \{x \in X \mid g(x) = 1\} = \operatorname{supp}(f) \cap \operatorname{supp}(g). \end{align}
Let $1 \!-\! f \colon X \to \{0,1\}$ be the function defined by $(1 \!-\! f)(x) = 1 - f(x)$ for all $x \in X$. The support of $1 \!-\!f$ is the complement of the support of $f$. Indeed, by definition of support, \begin{align} \operatorname{supp}(1 \!-\! f) &= \{x \in X \mid f(x) = 0\} \\ &= \{x \in X \mid f(x) \neq 1\} \\ &= X \smallsetminus \{x \in X \mid f(x) = 1\} \ = X \smallsetminus \operatorname{supp}(f). \end{align}