Prove that A-(BnC)=(A-B)U (A-C)

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Solution

A-(B ⋂ C)=(A-B) ⋃ (A-C)

According to definition,

A-B={x| x ∈A and x∉B}

A-C={x |x ∈A and x∉ C}

then
(A-B) ⋃ (A-C)={x|x∈A and x∉(B and C)

Let X=A and Y=(B ⋂ C)
then
X-Y={x | x∈X and x∉Y}
x∉Y
x∉(B ⋂ C)
x∉(B and C)

A-(B ⋂ C) or X-Y = {x|x∈A and x∉(B and C)

Therefore,
A-(B ⋂ C)=(A-B)⋃(A-C)

Hence the proof

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