:-) $\begingroup$ I suppose if x ∈ (A ∩ C) then x ∈ A, which is an option given by A ∪ (B ∩ C), so I have not discovered a location in (A ∪ B) ∩ (A ∪ C) that does not hold for A ∪ (B ∩ C) as well. However, it feels significant that I can say x ∈ (A ∩ C) when looking at (A ∪ B) ∩ (A ∪ C), but not when looking at A ... B.C.), minister and kinsman of a petty kinglet under the Chou dynasty, whose `Li Sao', literally translated `Falling into Trouble', is partly autobiography and partly imagination.
View in context He is, by an almost universal consent, allowed to have been born about the year 620 B.C. , and to have been by birth a slave. b+c 6= a b + a c (b) Cancellation of the c here requires that it appears in each additive term of the numerator and denominator: ca+cb cd = c(a+b) cd = a+b d but ca+b cd 6= a+b d (c) Compound fractions can be simpliﬁed by using the rule "division is the same as Litery h 1, h 2 i h 3 oznaczają długości wysokości tego trójkąta poprowadzone odpowiednio do boków a, b i c.