Ja, deine Lösung ist richtig:
\( C \backslash[C \backslash[B \backslash(B \backslash A)]] \)
= \( C \backslash[C \backslash[B \backslash(B \cap \bar{A})]] \)
= \( C \backslash[C \backslash[B \cap \overline{B \cap \bar{A} }]] \)
= \( C \backslash[C \backslash[B \cap (\bar{B} \cup A ) ] ] \)
= \( C \backslash[C \backslash[(B \cap \bar{B}) \cup (B \cap A ) ] ] \)
= \( C \backslash[C \backslash[ B \cap A ] ] \)
etc.
