TUGAS 5 PDM

Show that A ⊂ B ⇔ A ∩ B = A
Proof:
(i) Show that A ⊂ B ⇒ A ∩ B = A
We have A ⊂ B
it means that ∀ x ∈ A, x ∈ B
show that A ⊂ (A ∩ B)
take any x ∈ A
obvious x ∈ A ∧ x ∈ A
⇔ x ∈ A ∧ x ∈ B (because ∀ x ∈ A, x ∈ B)
⇔ x ∈ (A ∩ B)
We get for ∀ x ∈ A then x ∈ (A ∩ B)
it means A ⊂ (A ∩ B)....................(1)
take any x ∈ (A ∩ B)
show that (A ∩ B) ⊂ A
obvious x ∈ (A ∩ B)
⇔ x ∈ A ∧ x ∈ B (simplifikasi)
⇔ x ∈ A
we get for all x ∈ (A ∩ B), x ∈ A
it means (A ∩ B) ⊂ A....................(2)
From (1) and (2) we conclude that A ∩ B=A

So, if A ⊂ B then A ∩ B = A

Versi kontradiksi:
Take any x ∈ (A ∩ B)
Show that (A ∩ B) ⊂ A
andaikan(A ∩ B) ⊄ A maka∃ x ∈ (A ∩ B), akan tetapi x ∉ A
maka x ∈ A ∩ B ∧ x ∈ Ac
⇔ (x ∈ A ∧ x ∈ B) ∧ x ∈ Ac (assosiatif)
⇔ (x ∈ A ∧ x ∈ Ac) ∧ x ∈ B
⇔ x ∈ (A ∧ Ac) ∧ x ∈ B
⇔ x ∈ ∅ ∧ x∈ B
⇔ x ∈ (∅ ∩ B)
⇔ x ∈ ∅
maka terjadi kontradiksi oleh sebab terdapat x ∈ (A ∩ B) dan x ∉ A berlaku x ∈ ∅
Jadi pengandaian ditolak
Jadi yang benar adalah (A ∩ B) ⊂ A

(ii) Show that A ∩ B = A ⇒ A ⊂ B
We have A ∩ B = A, it means (A ∩ B) ⊂A and A ⊂ (A ∩ B)
Show that (A ∩ B) ⊂A
take any x ∈ (A ∩ B)
Obvious x ∈ A ∧ x ∈ B (simplifikasi)
⇔ x ∈ A
We get for all x ∈ (A ∩ B) then x ∈ A
it means that (A ∩ B) ⊂ A..................(1)
take any x ∈ A
obvious x ∈ A ∧ x ∈ A
⇔ x∈ A ∧ x ∈ B (because ∀ x ∈ A, x ∈ B)
⇔ x∈ (A ∩ B)
We get for all x ∈ A then x ∈ (A ∩ B)
it means that A ⊂ (A ∩ B)...................(2)
From (1) and (2) we conclude that A ∩ B = A

Because in rule (2) shown that x ∈ A, x ∈ B so we conclude A ⊂ B

So, if A ∩ B = A then A ⊂ B

Final conclusion : A ⊂ B ⇔ A ∩ B = A