\( X \backslash\left(M_{1} \cap M_{2}\right)=\left(X \backslash M_{1}\right) \cup\left(X \backslash M_{2}\right) \)
Zeige die Gleichung - wie üblich bei Mengengleichheit - etwa so:
Sei x∈   \( X \backslash\left(M_{1} \cap M_{2}\right) \)
==>  x∈X  ∧  x∉ \( M_{1} \cap M_{2} \)
==>  x∈X  ∧  (  x∉ \( M_{1}  \)  ∨  x∉  \( M_{2} \) )
distributiv bei ∧ und  ∨
==>  (x∈X  ∧  x∉ \( M_{1}  \)  )  ∨  (x∈X  ∧ x∉  \( M_{2} \) )
=  \( \left(X \backslash M_{1}\right) \cup\left(X \backslash M_{2}\right) \)
Damit hast du
\( X \backslash\left(M_{1} \cap M_{2}\right)⊆\left(X \backslash M_{1}\right) \cup\left(X \backslash M_{2}\right) \)
Die andere Inklusion entsprechend.