Answer:
[tex]A \cup (C-B) = (A \cup C) \cap (A \cup B^c)\\A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \,\,\, \text{Here you use double inclusion.}\\A \cup B \cup C = A \cup (B \cup C)\\B^c \cap (A-C) = (B^c - C) \cap A\\(B \cap C) - A = (B-A)\cap C\\C-(A\cup B) = (C - A )\cap B\\A^c \cap (B \cap C) = ( A^c \cap B) \cap C[/tex]
Step-by-step explanation:
[tex]A \cup (C-B) = A \cup (C \cap B^c) = (A \cup C) \cap (A \cup B^c)\\A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \,\,\, \text{Here you use double inclusion.}\\A \cup B \cup C = (A \cup B ) \cup C = A \cup (B \cup C)\\B^c \cap (A-C) = B^c \cap (A \cap C^c) = (B^c \cap C^c) \cap A = (B^c - C) \cap A\\(B \cap C) - A = (B \cap C) \cap A^c = (B \cap C) \cap A^c = (B \cap A^c)\cap C = (B-A)\cap C\\C-(A\cup B) = C \cap (A\cup B)^c = C \cap (A^c \cap B^c ) = (C \cap A^c )\cap B^c = (C - A )\cap B\\[/tex]
[tex]A^c \cap (B \cap C) = ( A^c \cap B) \cap C[/tex]