Fuzzy Intersection is Commutative

Theorem

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:

$B \cap A$

By Intersection is Commutative:

$A \cap B = B \cap A$

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.

$\blacksquare$