Show that An(B-C)=(AnB)-(AnC)
1) Let x be an arbitrary element in A (B-C)
x A and x (B-C)
x A, and x B and x C
In either case, this implies x (A and B) but x (A and C)
x (AB) - (AC)
Hence, A (B-C) (AB) - (AC) ------- (1)
2) Again let y be an arbitrary element in (AB) - (AC)
y (AB) but y (AC)
y A, and y B but y A and y C
y A (B-C)
So, (AB) - (AC) A (B-C) ---- (2)
Thus from (1) & (2), it is proved that
A (B-C) = (AB) - (AC)
x A and x (B-C)
x A, and x B and x C
In either case, this implies x (A and B) but x (A and C)
x (AB) - (AC)
Hence, A (B-C) (AB) - (AC) ------- (1)
2) Again let y be an arbitrary element in (AB) - (AC)
y (AB) but y (AC)
y A, and y B but y A and y C
y A (B-C)
So, (AB) - (AC) A (B-C) ---- (2)
Thus from (1) & (2), it is proved that
A (B-C) = (AB) - (AC)