⇔
A ∩ B = AProof:
(i) Show that A ⊂ B
⇒
A ∩ B = AWe have A ⊂ B
it means that
∀
x ∈ A, x ∈ Bshow 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 ∈ Awe 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 ∉ Amaka
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 ⊂ BWe 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 ⊂ BFinal conclusion : A ⊂ B
⇔
A ∩ B = A