Question

(AnB) -c=(A-C) n(B-C) how can we prove it
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Answer 1

$\large\underline{\sf{Solution-}}$

Consider,

$\sf \: A\cap (B - C)$

$\sf \: = \: A\cap (B\cap C')$

can be further rewritten as

$\sf \: = \: \red{ \phi} \: \cup \: [ A\cap (B\cap C') ]$

can be further rewritten as

$\sf \: = \: \red{[ (A\cap B)\cap A']} \: \cup \: [ (A\cap B)\cap C']$

$\sf \: = \: (A\cap B)\cap (A'\cup C')$

$\sf \: = \: (A\cap B)\cap (A\cap C)'$

$\sf \: = \: (A\cap B) - (A\cap C)$

Hence,

$\implies\sf \: \boxed{\bf \: A\cap (B - C) = (A\cap B) - (A\cap C) \: }$

$\rule{190pt}{2pt}$

$\sf \: \large Law\:used-$

$\sf \: \phi \: \cup \: A \: = \: A$

$\sf \: A - B = A\cap B'$

$\sf \: A\cap A' = \phi$

$\sf \: (A\cap B)\cap C = A\cap (B\cap C)$

$\sf \: A\cap (B\cup C) = (A\cap B)\cup (A\cap C)$

$\sf \: (A\cup B)' = A'\cap B'$