Question

4. Let A, B, and C be sets. Prove that AU(BNC) = (AUB) n (AUC).

Answer 1

this using mdual inclusion Sin We Pree prove (ADB)n (AUC) are Au CBnc) shew that First we LE We on elBnc xEA det xE AU (Bnc) Jhen Consider two Caser Care gir LEA xe CAUC) So xe CAUB) n(AUC) % CAUB) a nd Jhen A Case 2 xE BnC Since xeB we have that Jhen zEB and xEC %6 CAUC) has have that Xe CAUB) inie LEC e e (AUC)n CAU B) Le CAUB and xE (AUc Htnce Au (BAC) C (AUB) nCAUC) Jhws show that (AUB) n CAUC) C AU(Bnc) Next we e CAUB) and E (AUC) Xe CAUB) n(Au c)Jhen det Co nsider two Cases We Case 1 2EA Jhin ze (AU CBnc)) Case 2 xA But we know that fince 6 AUB we LE A LE B of that 14A So we must have that are LEB

(Bul we Rnow that E A or xEC xe AUC we sine that x4A f must have that we are LE C Jhus CAUB) n CAUC) AUCBnc) (AUB) n CAUC) c finte Aulenc) CAUB) n (AUc) and AnCBnc), aet Au Cenc) = CAUB)n CAuc) we