12 Oktober 2009

TUGAS 5 PDM

UNION

1. Show that (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)
Proof:
i) Show that (A ∪ B) ∩ C ⊂ (A ∩ C) ∪ (B ∩ C)
Take any x ∈ (A ∪ B) ∩ C
Obvious x ∈ (A ∪ B) ∩ C
≡ x ∈ (A ∪ B) ∧ x ∈ C
≡ (x ∈ A ∨ x ∈ B) ∧ x ∈ C
≡ (x ∈ A ∧ x ∈ C) ∨ (x ∈ B ∧ x ∈ C)
≡ x ∈ (A ∩ C) ∪ x ∈ (B ∩ C)
≡ x ∈ (A ∩ C) ∪ (B ∩ C) .......1)
We get for all x ∈ (A ∪ B) ∩ C then x ∈ (A ∩ C) ∪ (B ∩ C).
It means (A ∪ B) ∩ C ⊂ (A ∩ C) ∪ (B ∩ C)

ii) Show that (A ∩ C) ∪ (B ∩ C) ⊂ (A ∪ B) ∩ C
Take any x ∈ (A ∩ C) ∪ (B ∩ C)
Obvious x ∈ (A ∩ C) ∪ (B ∩ C)
≡ x ∈ (A ∩ C) ∨ (B ∩ C)
≡ (x ∈ A ∧ x ∈ C) ∨ (x ∈ B ∧ x ∈ C)
≡ (x ∈ A ∧ x ∈ C) ∨ (x ∈ B ∧ x ∈ C)
≡ x ∈ C ∧ (x ∈ A ∨ x ∈ B) (Asso)
≡ (x ∈ A ∨ x ∈ B) ∧ x ∈ C
≡ x ∈ (A ∪ B) ∩ C .......2)
From 1) and 2) we conclude that (A ∪ B) ∩ C is true.
So, (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)

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