# Fuzzy Intersection is Commutative

## Proof

Let $\textbf A = \struct{A, \mu_A}$ and $\textbf B = \struct{B, \mu_B}$ be fuzzy sets.

### Proving Domain Equality

By the definition of fuzzy intersection the domain of $\textbf A \cap \textbf B$ is:

$A \cap B$

Similarly the domain of $\textbf B \cap \textbf A$ is:

$B \cap A$
$A \cap B = B \cap A$

Hence their domains are equal.

$\Box$

### Proving Membership Function Equality

#### Proving Form Equality

By the definition of fuzzy intersection the membership function of $\textbf A \cap \textbf B$ is of the form:

$\mu:A \cap B \to \closedint {0} {1}$

Similarly, the membership function of $\textbf B \cap \textbf A$ is of the form:

$\mu:B \cap A \to \closedint {0} {1}$

By Intersection is Commutative this is the same as:

$\mu:A \cap B \to \closedint {0} {1}$

Hence the membership functions are of the same form.

#### Proving Rule Equality

 $\ds \forall x \in A \cap B: \,$ $\ds \map {\mu_{A \mathop \cap B} } x$ $=$ $\ds \map \min {\map {\mu_A} x, \map {\mu_B} x}$ $\ds \leadstoandfrom \ \$ $\ds \forall x \in B \cap A: \,$ $\ds \map {\mu_{B \mathop \cap A} } x$ $=$ $\ds \map \min {\map {\mu_A} x, \map {\mu_B} x}$ Intersection is Commutative $\ds \leadstoandfrom \ \$ $\ds \forall x \in B \cap A: \,$ $\ds \map {\mu_{B \mathop \cap A} } x$ $=$ $\ds \map \min {\map {\mu_B} x, \map {\mu_A} x}$ Min Operation is Commutative

Hence the membership functions have the same rule.

$\blacksquare$