what is the proof for this?
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A few useful properties that I assume you know:
(1) x∈A∪B means x∈A or x∈B
(2) x∈A∩B means x∈A and x∈B
(3) If x∈A, then x∈A∪B for any set B (follows from property 1)
(4) A⊆B means that for any x∈A, we have x∈B.
Suppose x ∈(P∪Q)∩R. Then x∈P∪Q and x ∈R (by property 2 above). So, x∈R, and either x∈P or x∈Q (by property 1). If x∈P, then x∈P∪(Q∩R) (by property 3). If x∈Q, then x∈R and x∈Q, meaning x∈Q∩R (by property 2), so x∈P∪(Q∩R) (by property 3). Thus, for any x∈(P∪Q)∩R, we have x∈P∪(Q∩R), and so we conclude that (P∪Q)∩R⊆P∪(Q∩R) (by property 4).
Hope that helps :)
(1) x∈A∪B means x∈A or x∈B
(2) x∈A∩B means x∈A and x∈B
(3) If x∈A, then x∈A∪B for any set B (follows from property 1)
(4) A⊆B means that for any x∈A, we have x∈B.
Suppose x ∈(P∪Q)∩R. Then x∈P∪Q and x ∈R (by property 2 above). So, x∈R, and either x∈P or x∈Q (by property 1). If x∈P, then x∈P∪(Q∩R) (by property 3). If x∈Q, then x∈R and x∈Q, meaning x∈Q∩R (by property 2), so x∈P∪(Q∩R) (by property 3). Thus, for any x∈(P∪Q)∩R, we have x∈P∪(Q∩R), and so we conclude that (P∪Q)∩R⊆P∪(Q∩R) (by property 4).
Hope that helps :)
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I'm sorry about my answer earlier, i posted the answer to a different equation by accident. Forgive me.