Pengantar Dasar Matematika

**Task 1**

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)

     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)

      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)

**Task 2**

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 )