Tugas PDM 5

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 (distributif)
⇔ (X ∈ A∧ X ∈ C) ∨ (X ∈ B ∧ X ∈ C)
⇔ X ∈ ( A ∩C ) ∨ X ∈ (B∩C)
⇔ X ∈ ( A ∩C ) ∪ ( B∩C)

We get for All X ∈ ( A ∪ B ) ∩ C then X ∈ ( A ∩C ) ∪ ( B∩C)
It means
(A ∪ B ) ∩ C ⊂ ( A ∩ C ) ∪ ( B∩C) . . . . . . . . . . . . . (1)

With the same way get (2)
From (1) and (2) we conclude that ( A ∪ B ) ∩ C = ( A ∩ C ) ∪ ( B∩ C) is true.
So, ( A ∪ B ) ∩ C = ( A ∩ C ) ∪ ( B ∩ C)



Show that A ⊂ B if and only if ( A ∩ B ) = A
Proof :
(i)Show that A ⊂ B ⇒ (A ∩ B ) = A
We have A ⊂ B
It means ∀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 all 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
⇔ X ∈ A ( Simplifikasi )
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


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 ∈ ∅ Λ 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