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For all sets a,B and C show that `(A-B)nn(A-C)=A-(BuuC)` |
Answer» Let `x in (A-B) nn (A-C)` `rArr x in (A-B)` and `x in (A-C)` ` rArr (x in A "and" x !in B)` and (`x in A` and `x !in C`) `rArr x in A` and `(x !in B "and" x !in C)` `rArr x in A` and `x !in (B uu C)` `rArr x in A - (B uu C)` `rArr (A-B) nn (A - C) sub A - (B uu C)"............"(i)` Now, let ` y in A - (B uu C)` `rArr y in A` and ` y !in (B uu C)` `rArr y in A` and `y !in (B uu C)` `rArr y in A` and `(y !in B "and" y !in C)` `rArr (y in A and y !in B) "and" (y in A "and" y !in C)` `rArr y in (A - B)` and `y in (A - C)` `rArr y in (A - B) nn (A - C)` `rArr A - (B uu C) sub (A - B) nn (A - C) "......"(ii)` From Eqs. (i) and (ii) `A - (B uu C) = (A - B) nn (A - C)` |
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