Tugas 5

1) Show that (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C)
proof :
(i) Show that (A ∩ B) ∪ C ⊂ (A ∪ C) ∩ (B ∪ C)
Take any Χ ∈ (A ∩ B) ∪ C
Obvious Χ ∈ (A ∩ B) ∪ C
↔ Χ ∈ (A ∩ B) ∨ Χ ∈ C
↔ Χ ∈ A ∧ Χ∈ B ∨ Χ ∈ C (Distributif)
↔ Χ ∈ A ∨ Χ∈ C ∧ Χ ∈ B ∨ Χ ∈ C
↔ Χ ∈ (A ∪ C) ∧ Χ ∈ (B ∪ C)
↔ Χ ∈(A ∪ C) ∩ (B ∪ C)
We get for all Χ ∈ (A ∩ B) ∪ C that Χ ∈ (A ∪ C) ∩ (B ∪ C) It is means
(A ∩ B) ∪ C ⊂ (A ∪ C) ∩ (B ∪ C)....(1)

(ii)Show that (A ∩ B) ∪ C ⊂ (A ∪ C) ∩ (B ∪ C)
Take any Χ ∈ (A ∪ C) ∩ (B ∪ C)
Obvious Χ ∈ (A ∪ C) ∩ (B ∪ C)
↔ Χ ∈ (A ∪ C) ∧ Χ ∈ (B ∪ C)
↔ Χ ∈ A ∨ Χ ∈ C ∧ Χ ∈ B ∨ Χ ∈ C
↔ Χ ∈ (A ∧ B)∨ Χ ∈ C
↔ Χ ∈ (A ∩ B)∪ Χ ∈ C
↔ Χ ∈ (A ∩ B) ∪ C....(2)
So (A ∩ B) ∪ C ⊂ (A ∩ C) ∩ (B ∪ C)
From (1) and (2) we conclude that (A ∩ B) ∪ C = (A ∪ C) ∪ (B ∪ C)

2) Show that (A ∪ B)∩ C = (A ∩ C) ∪ (B ∩ C)
proof :
(i) Show that (A ∪ B) ∩ C ⊂ (A ∩ C) ∪ (B ∩ C)
Take any Χ ∈ (A ∪ B) ∩ C
Obvious Χ ∈ (A ∪ B) ∩ C
↔ Χ ∈ (A ∪ B) ∧ Χ ∈ C
↔ Χ ∈ A ∨ Χ∈ B ∧ Χ ∈ C (Distributif)
↔ Χ ∈ A ∧ Χ∈ C ∨ Χ ∈ B ∨ Χ ∈ C
↔ Χ ∈ (A ∩ C) ∨ Χ ∈ (B ∩ C)
↔ Χ ∈(A ∩ C) ∪ (B ∩ C)
We get for all Χ ∈ (A ∪ B) ∩ C that Χ ∈ (A ∩ C) ∪ (B ∩ C) It is means
(A ∪ B) ∩ C ⊂ (A ∩ C) ∪ (B ∩ C)....(1)

(ii)Show that (A ∪ B) ∩ C ⊂ (A ∩ C) ∪ (B ∩ C)
Take any Χ ∈ (A ∩ C) ∪ (B ∩ C)
Obvious Χ ∈ (A ∩ C) ∪ (B ∩ C)
↔ Χ ∈ (A ∩ C) ∨ Χ ∈ (B ∩ C)
↔ Χ ∈ A ∧ Χ ∈ C ∨ Χ ∈ B ∧ Χ ∈ C
↔ Χ ∈ (A ∨ B)∧ Χ ∈ C
↔ Χ ∈ (A ∪ B) ∩ C....(2)
So (A ∪ B) ∩ C ⊂ (A ∩ C) ∪ (B ∩ C)
From (1) and (2) we conclude that (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)