I have been studying Fuzzy Set Theory and in that, we come across standard fuzzy union ($\cup$) and intersection ($\cap$) defined as the maximum and minimum of membership values respectively.
In some proof regarding distributive laws, I came across the expression,
$$\min \left\lbrace A(x), \max \left\lbrace B(x), C(x) \right\rbrace \right\rbrace$$ where these $A(x), B(x) \text{ and } C(x)$ are membership values of $x$ in $\left[ 0, 1 \right]$.
However, to prove the distribution of intersection over addition, i.e., $A \cap \left( B \cup C \right) = \left( A \cap B \right) \cup \left( A \cap C \right)$, I will need to prove
$$\min \left\lbrace A(x), \max \left\lbrace B(x), C(x) \right\rbrace \right\rbrace = \max \left\lbrace \min \left\lbrace A(x), B(x) \right\rbrace, \min \left\lbrace A(x), C(x) \right\rbrace \right\rbrace$$.
Any insights about how to prove this will be useful!
If $A(x)$ is less than or equal to both $B(x)$ and $C(x)$, then the right-hand side will evaluate to $\max\{A(x), A(x)\} = A(x)$, the same as the left-hand side.
If $A(x)$ is greater than either one of $B(x)$ and $C(x)$, say $B(x)$, but not the other (so that $A(x) \leq \max\{B(x), C(x)\}$ still), then the right-hand side will be $\max \{B(x), A(x)\} = A(x)$, again the same as the left-hand side.
Finally, if $A(x)$ is greater than both $B(x)$ and $C(x)$, then the right-hand side will be $\max\{B(x),C(x)\}$, which yet again is the same as the left-hand side in this case. This finishes the proof.
Basically, case-checking alone will suffice to show the proof; there may be a more elegant way, but this certainly gets the job done.