Respuesta :

discipulus

Answer:

[tex] let \:x \in (A\cap B)'\\\Rightarrow x \notin (A\cap B) \\\Rightarrow x \notin A \:or\: x\notin B\\ \Rightarrow x \in A' \:or \:x \in B'\\\Rightarrow x \in A' \cup B'\\\therefore (A ∩ B)' = A' ∪ B' [/tex]