Prove that two fuzzy sets are disjoint if and only if their supports are disjoint.
Given two fuzzy sets $A,B$ of a reference set $X$,then :
$$ \begin{align} \\ &\text{Supp}(A) \cap \text{Supp}(B)= \emptyset\\ &\iff \nexists x \in X:x \in \text{Supp}(A) \wedge x \in \text{Supp}(B)\tag1\\ &\iff \nexists x \in X:\mu_A(x)>0 \wedge \mu_B(x)>0\tag2\\ &\iff A \cap B=\emptyset \tag3 \end{align} $$
$\hphantom{0}(1)$ Definition of $\cap$
$\hphantom{0}(2)$ Definition of $\text{Supp}$
$\hphantom{0}(3)$ Definition of disjoint fuzzy sets
The other way I came up with was:
$$\text{Supp}(A)=\{x \in X : \mu_{A}(x) >0\}=\{x \in X : x \in A\}=A \tag{I}$$
Hence if $A\cap B=\emptyset $ then it's enough to set $\text{Supp}(A)=A$ and $\text{Supp}(B)=B$, the other direction is the same, and then we conclude the result.
I want to know that how much of my work is correct, I think the first one is correct, but about the other one I doubt if for all fuzzy sets $(\text{I})$ does hold.
$\DeclareMathOperator{Supp}{Supp}$(I) does not make sense. $A$ is a fuzzy set, while $\Supp(A)$ is a set. So saying $\Supp(A) = A$ doesn't "type check".
Your proof is actually almost correct. Here is the complete proof.
$\begin{equation} \begin{split} \Supp(A) \cap \Supp(B) = \emptyset &\iff \forall x \in X . \neg (x \in \Supp (A) \land x \in \Supp (B)) \\ &\iff \forall x \in X . \neg (\mu_A(x) > 0 \land \mu_B(x) > 0) \\ &\iff \forall x \in X . \min(\mu_A(x), \mu_B(x)) \leq 0 \\ &\iff \forall x \in X . \min(\mu_A(x), \mu_B(x)) = 0 \\ &\iff \forall x \in X . \mu_{A \cap B}(x) = 0 \\ &\iff \forall x \in X . \mu_{A \cap B}(x) = \mu_\emptyset(x) \\ &\iff A \cap B = \emptyset \end{split} \end{equation} $
It's easier to do the proof using $\forall x \in X$ rather than $\nexists x \in X$.
Your proof is correct that $\Supp(A) \cap \Supp(B) = \emptyset \iff (1) \iff (2)$. But you did not actually prove that $(2) \iff A \cap B = \emptyset$. This is the gap which my proof fills in.