Proof A⊂B if and only if A∩B = A

Show that A⊂B if and only if A∩B = A!
Proof :
Show that (A⊂B) ↔ (A∩B = A)
we have A⊂B,
it means x∈A, x∈B.

i. Show that A∩B ⊂ A
Take any x∈ A∩B
Obvious x∈A ∧ x∈B
↔ x∈A (simplifikasi)
We get for all x∈ A∩B, then x∈A
It means A∩B ⊂ A.......................1)

ii. Show that A ⊂ A∩B
Take any x∈ A
Obvious x∈A ∧ x∈A
↔ x∈A ∧ x∈B (because A⊂B,∀ x∈A, x∈B )
↔ x∈(A∩B)
We get for all x∈A, then x∈ A∩B
It means A ⊂ A∩B.......................2)

From (1) and (2) we conclude that A∩B = A is true,
So, if A⊂B then A∩B = A

TUGAS PDM 4

Show that :
a) A ∩ A = A
b) A ∩ B = B ∩ A
c) ( A ∩ B ) ∩ C = A ∩ ( B ∩ C )

Answer

a) Proof :

i. Show that A ∩ A ⊂ A
Take any x ∈ A ∩ A
Obvious x ∈ A ∩ A
≡ x ∈ A ∧ x ∈ A
≡ x ∈ A (idempoten)
So, A ∩ A ⊂ A .................................(1)

ii. Show that A ⊂ A ∩ A
Take any x ∈ A
Obvious x ∈ A
≡ x ∈ A
≡ x ∈ A ∧ x ∈ A (idempoten)
So, A ⊂ A ∩ A .................................(2)

From (1) and (2), we conclude that A ∩ A = A


b) Proof :

i. Show that ( A ∩ B ) ⊂ ( B ∩ A )
Take any x ∈ A ∩ B
Obvious x ∈ A ∩ B
≡ x ∈ A ∧ x ∈ B
≡ x ∈ B ∧ x ∈ A (komutatif)
So, ( A ∩ B ) ⊂ ( B ∩ A )......................(1)

ii. Show that( B ∩ A ) ⊂ ( A ∩ B )
Take any x ∈ B ∩ A
Obvious x ∈ B ∩ A
≡ x ∈ B ∧ x ∈ A
≡ x ∈ A ∧ x ∈ B (komutatif)
So, ( B ∩ A ) ⊂ ( A ∩ B ) ......................(2)

From (1) and (2), we conclude that A ∩ B = B ∩ A

c) Proof :

i. Show that ( A ∩ B ) ∩ C = A ∩ ( B ∩ C )
Take any x ∈ ( A ∩ B ) ∩ C
Obvious x ∈ ( A ∩ B ) ∩ C
≡ (x ∈ A ∧ x ∈ B) ∧ x ∈ C
≡ x ∈ A ∧ ( x ∈ B ∧ x ∈ C ) (asosiatif)
So, [( A ∩ B ) ∩ C] ⊂ [A ∩ ( B ∩ C )]...........(1)

ii. Show that A ∩ ( B ∩ C ) = (A ∩ B ) ∩ C
Take any x ∈ A ∩ ( B ∩ C )
Obvious x ∈ A ∩ ( B ∩ C )
≡ x ∈ A ∧ ( x ∈ B ∧ x ∈ C)
≡ ( x ∈ A ∧ x ∈ B ) ∧ x ∈ C (asosiatif)
So, [A ∩ ( B ∩ C )] ⊂ [( A ∩ B ) ∩ C] ...........(2)

From (1) and (2), we conclude that ( A ∩ B ) ∩ C = A ∩ ( B ∩ C )

TUGAS PDM 3

4.a) 1. [(a ∨ c) ∧ ~b] → [(d→ c) → f]
2. ~a→ b
3. ~b /∴ [(d→ c) → f]
4. a (2,3 MT)
5. a ∨ c (4 add)
6. (a ∨ c) ∧ ~b (5,3 konj)
7. (d→ c) → f (1,6 MP)

4.b) 1. e→ (f ∧~g)
2. (f ∨ g)→ h
3. e /∴ h
4. (f ∧~g) (1,3 MP)
5. f (4 simp)
6. (f ∨ g) (5 add)
7. h (2,6 MP)

4.c) 1. e→ f
2. e→ g /∴ e→ (f ∧ g)
3. (e→ f) ∧ (e→ g) (1,2 konj)
4. e→ (f ∧ g) (3 dist)

4.d) 1. (~u ∨ v) ∧ (u ∨ v)
2. ~x→ ~w /∴ v ∨ x
3. (~u ∧ u) ∨ v (1 dist)
4. F ∨ v (3 komp)
5. v (4 id)
6. v ∨ x (5 add)

4.e) 1. e→ f
2. g→ f /∴(e ∨ g) → f
3. ~f→ ~e (1 ekiv)
4. ~f→ ~g (2 ekiv)
5. (~f→ ~e) ∧ (~f→ ~g) (3,4 konj)
6. ~f→ (~e ∧ ~g) (5 dist)
7. (e ∨ g) → f (6 ekiv)

5.a) 1. b → n
2. ~b → s / ∴ n ∨ s
3. ~n → ~b (1 ekiv)
4. ~n → s (3,2 sil)
5. n ∨ s (4 ekiv)

5.b) 1. ( p ∧ t) → n
2. ( t → n) → s
3. p / ∴ s
4. p → (t → n) (1 eksp)
5. t → n (4,3 MP)
6. s (2,5 MP)

5.d) 1. b ∨ k
2. ( b ∨ m ) → ( l ∧ h )
3. ~l / ∴ k
4. ( ~l ∨ ~h) → ( ~b ∧ ~m) (2 ekiv)
5. ~l ∨ ~h (3 add)
6. ~b ∧ ~m (4,5 MP)
7. ~b (6 simp)
8. ~b → k (1 ekiv)
9. k (8,9 MP)