Fuzzy Intersection is Commutative
Jump to navigation
Jump to search
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.
$\blacksquare$