Question

if A={a,b,d,e,f}
B={d,e,f,g,h}
C=b,{b,c,d,e}
verify Au(BnC)=(AuB)n(AuC)​

Answer 1

We have,

A = { a , b , d , e , f }

B = { d , e , f , g , h }

C = { b , c , d , e }

$\tt expression \: : \: A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$

$\tt l.h.s \\ \\ = \sf A \cup (B \cap C) \\ \\ = \sf \{a, \: b, \: d, \: e, \: f \} \cup \left( \{ d, \: e, \: f, \: g, \: h \} \cap \{b, \: c, \: d, \: e \} \right) \\ \\ = \sf \{a, \: b, \: d, \: e, \: f \} \cup \{d, e\} \\ \\ \sf = \{ a, \: b, \: d, \: e, \: f \}$

$\tt r.h.s \\ \\ = \sf (A \cup B) \cap (A \cup C) \\ \\ = \sf ( \{a, \: b, \: d, \: e, \: f \} \cup \{ d, \: e, \: f, \: g, \: h \}) \cap \left( \{ a, \: b, \: d, \: e, \: f \} \cup \{b, \: c, \: d, \: e \} \right) \\ \\ = \sf \{a, \: b, \: d, \: e, \: f, \: g, \: h, \} \cap \{a, \: b, \: c, \: d, \: e, \: f \} \\ \\ \sf = \{ a, \: b, \: d, \: e, \: f \}$

Hence, L.H.S = R.H.S