Question

An[B-C)=(AnB)-(AnC)
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Answer 1

Answer:

proved

Step-by-step explanation:

Given to prove An[B-C)=(A n B) - (A n C)

Let p be an arbitrary element in A ∩ (B – C)

p Є A and p Є (B – C)

implies p Є A, p Є B and p does not belong to C

Implies p Є A and B but p does not belong to A and C

It implies p Є (A ∩ B) – (A ∩ C)

So, A ∩ (B – C) is a subset of (A ∩ B) – (A ∩ C) -----------(1)

Now let q be an arbitrary element in (A ∩B) – (A ∩ C)

q belongs to (A ∩ B) but q does not belong to (A ∩ C)

Implies q Є A and q Є B but q Є A and q does not belong to C

Implies q Є A ∩ (B – C)

Now (A ∩ B) – (A ∩ C) is a subset of A ∩ (B – C) -----------(2)  

Now from 1 and 2 we get

A ∩ (B – C) = (A ∩ B) – (A ∩ C)

Answer 2

Answer:

Step-by-step explanation: