Tugas 5

1) Show that (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C)
proof :
(i) Show that (A ∩ B) ∪ C ⊂ (A ∪ C) ∩ (B ∪ C)
Take any Χ ∈ (A ∩ B) ∪ C
Obvious Χ ∈ (A ∩ B) ∪ C
↔ Χ ∈ (A ∩ B) ∨ Χ ∈ C
↔ Χ ∈ A ∧ Χ∈ B ∨ Χ ∈ C (Distributif)
↔ Χ ∈ A ∨ Χ∈ C ∧ Χ ∈ B ∨ Χ ∈ C
↔ Χ ∈ (A ∪ C) ∧ Χ ∈ (B ∪ C)
↔ Χ ∈(A ∪ C) ∩ (B ∪ C)
We get for all Χ ∈ (A ∩ B) ∪ C that Χ ∈ (A ∪ C) ∩ (B ∪ C) It is means
(A ∩ B) ∪ C ⊂ (A ∪ C) ∩ (B ∪ C)....(1)

(ii)Show that (A ∩ B) ∪ C ⊂ (A ∪ C) ∩ (B ∪ C)
Take any Χ ∈ (A ∪ C) ∩ (B ∪ C)
Obvious Χ ∈ (A ∪ C) ∩ (B ∪ C)
↔ Χ ∈ (A ∪ C) ∧ Χ ∈ (B ∪ C)
↔ Χ ∈ A ∨ Χ ∈ C ∧ Χ ∈ B ∨ Χ ∈ C
↔ Χ ∈ (A ∧ B)∨ Χ ∈ C
↔ Χ ∈ (A ∩ B)∪ Χ ∈ C
↔ Χ ∈ (A ∩ B) ∪ C....(2)
So (A ∩ B) ∪ C ⊂ (A ∩ C) ∩ (B ∪ C)
From (1) and (2) we conclude that (A ∩ B) ∪ C = (A ∪ C) ∪ (B ∪ C)

2) Show that (A ∪ B)∩ C = (A ∩ C) ∪ (B ∩ C)
proof :
(i) Show that (A ∪ B) ∩ C ⊂ (A ∩ C) ∪ (B ∩ C)
Take any Χ ∈ (A ∪ B) ∩ C
Obvious Χ ∈ (A ∪ B) ∩ C
↔ Χ ∈ (A ∪ B) ∧ Χ ∈ C
↔ Χ ∈ A ∨ Χ∈ B ∧ Χ ∈ C (Distributif)
↔ Χ ∈ A ∧ Χ∈ C ∨ Χ ∈ B ∨ Χ ∈ C
↔ Χ ∈ (A ∩ C) ∨ Χ ∈ (B ∩ C)
↔ Χ ∈(A ∩ C) ∪ (B ∩ C)
We get for all Χ ∈ (A ∪ B) ∩ C that Χ ∈ (A ∩ C) ∪ (B ∩ C) It is means
(A ∪ B) ∩ C ⊂ (A ∩ C) ∪ (B ∩ C)....(1)

(ii)Show that (A ∪ B) ∩ C ⊂ (A ∩ C) ∪ (B ∩ C)
Take any Χ ∈ (A ∩ C) ∪ (B ∩ C)
Obvious Χ ∈ (A ∩ C) ∪ (B ∩ C)
↔ Χ ∈ (A ∩ C) ∨ Χ ∈ (B ∩ C)
↔ Χ ∈ A ∧ Χ ∈ C ∨ Χ ∈ B ∧ Χ ∈ C
↔ Χ ∈ (A ∨ B)∧ Χ ∈ C
↔ Χ ∈ (A ∪ B) ∩ C....(2)
So (A ∪ B) ∩ C ⊂ (A ∩ C) ∪ (B ∩ C)
From (1) and (2) we conclude that (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)

TUGAS 4

1. Let A,B set and x is an elements. While :
a. x ∈ A ∩ B
b. x∉ A ∩ B

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

The answer:
1. a). x ∈ A ∩ B ↔ x ∈ A ∧ x ∈ B
b). PBE :
1. x ∉ A ∩ B
2. Tidak benar bahwa x ∈ A ∩ B
3. Tidak benar bahwa x ∈ A ∧ x ∈ B
4. x ∉ A ∧ x ∉ B

2. a). Proof :
i) Show that A ∩ A ⊂ A
Take any x ∈ A ∩ A
Obvious x ∈ A ∩ A
↔ x ∈ A ∧ x ∈ A
↔ x ∈ A
So, A ∩ A ⊂ A
ii) Show that A ⊂ A ∩ A
Take any x ∈ A
Obvious x ∈ A
↔ x ∈ A ∧ x ∈ A
↔ x ∈ A
So, A ⊂ A ∩ A
From (i) and (ii) we conclude that A ∩ A = A

b). Proof :
i). Show that A ∩ B ⊂ B ∩ A
Take any x ∈ A ∩ B ∧ x∈ B∩ A
Obvious x ∈ A ∩ B ∧ x ∈ B ∩ A
↔ x ∈(A ∩ B) ∧ x ∈ (B ∩ A)
↔ x ∈ {(A ∩ B) ∧ (B ∩ A)}
So A ∩ B ⊂ B ∩ A

ii). Show that B ∩A ⊂ A ∩ B
Take any x ∈ B ∩ A ∧ x ∈ A ∩ B
Obvious x ∈ B ∩ A ∧ x ∈ A ∩ B
↔ x ∈ (B ∩ A) ∧ x ∈ (A ∩ B)
↔ x ∈ {(B ∩ A) ∧ (A ∩ B)}
So B ∩ A ⊂ A ∩ B
From (i) and (ii) we conclude that A ∩ B = B ∩ A (H.Komutatif)

c). Proof :
i) Show that (A ∩ B)∩ C ⊂ A ∩ (B ∩ C)
Take any x ∈ (A ∩ B) ∩ C ∧ x ∈ A ∩ (B ∩ C)
Obvious x ∈ (A ∩ B) ∩ C ∧ x ∈ A ∩(B ∩ C)
↔ x ∈ (A ∩ B) ∩ C ∧ x ∈ A ∩ (B ∩ C)
↔ x ∈ {(A ∩ B) ∩ C ∧ A ∩ (B ∩ C)}
So (A ∩ B) ∩ C ⊂ A ∩ (B ∩ C)

ii) Show that A ∩(B ∩ C) ⊂ (A ∩ B)∩ C
Take any x ∈ A ∩(B ∩ C) ∧ x ∈ (A ∩ B)∩ C
Obvious x ∈ A ∩(B ∩ C) ∧ x ∈(A ∩ B)∩ C
↔ x ∈ A ∩ (B ∩ C) ∧ x ∈ (A ∩ B) ∩ C
↔ x ∈ {A ∩ (B ∩ C) ∧ (A ∩ B) ∩ C}
So A ∩(B ∩ C) ⊂ (A ∩ B)∩ C
From (i) and (ii) we conclude that A ∩(B ∩ C) = (A ∩ B)∩ C

TUGAS 3

1.Modus Ponens (MP)
p→q
p /∴ q

[(p→q) ∧ p]→q
≡ [(~p∨ q)∧ p]→q (imp)
≡ [(~p∧ p) ∨ (q ∧p)]→q (dist)
≡ [ F ∨ (q ∧p)]→q (komp)
≡ (q ∧p)→q (id)
≡ (~q ∨~p) ∨ q (imp)
≡ (~q∨ q) ∨ ~p (aso)
≡ T ∨~p (komp)
≡ T (id)


2. Modus Tollens (MT)
p→q
~q / ∴~p

[(p→q) ∧ ~q]→~p
≡ [(~p∨ q) ∧ ~q]→~p (imp)
≡ [(~p∧~q) ∨ (q ∧~q] →~p (dist)
≡ [(~p∧ ~q)∨ F] →~p (komp)
≡ (~p∧ ~q) →~p (id)
≡ (p ∨q) ∨ ~p (imp)
≡ (p ∨~p)∨ q (aso)
≡ T ∨q (komp)
≡ T (id)


3. Silogisme
p→q
q→r / ∴p→r

[(p→q) ∧ (q→r)]→( p→r)
≡ (p→q)→[(q→r)→( p→r)] (eksp)
≡ (p→q)→[(~q∨ r)→(~p∨ r)] (imp)
≡ (p→q)→[(q∧ ~r) ∨ (~p∨ r)] (imp)
≡ (p→q)→[(q ∧~r) ∨ (r∨ ~p)] (komp)
≡ (p→q)→[(q∧ ~r) ∨ r] ∨ ~p (aso)
≡ (p→q)→[(q∨ r)∧ (~r∨ r)] ∨~p (dist)
≡ (p→q)→[(q∨ r) ∧ T] ∨ p (komp)
≡ (p→q)→[(q∨ r) ∨ ~p (id)
≡ (~p∨ q)→q∨ r∨ ~p (imp)
≡ ~(~p∨ q) ∨ (q∨ r∨ ~p) (imp)
≡ ~(~p∨ q) ∨ (~p∧q) ∧ r (aso)
≡ T∨ r (komp)
≡ T (id)


4. Distruktif Silogisme (DS)
(p∨ q)
~p /∴ q

[(p ∨q) ∧ ~p]→q
≡ [(p∧ ~p) ∨ (q∧ ~p)]→q (dist)
≡ [F ∨ (q∧ ~p)]→q (komp)
≡ (q ∧~p) →q (id)
≡ (~q∨ p)∨ q (imp)
≡ (~q ∨q) ∨ p (aso)
≡ T∨ p (komp)
≡ T (id)


5.Konstruktif Dilema (KD)
p→q ∧(r→s)
(p ∨r) /∴ (q ∨s)

{[(p→q) ∧ (r→s)] ∧ (p∨ r)}→q∨ s)
≡ [(~p∨ q) ∧ (~r ∨s) ∧ (p∨ r)]→(q∨ s) (imp)
≡ [(p∧ ~q)∨ (r ∧~s)∨ (~p∧ ~r)]∨ (q)∨s) (imp)
≡ [(p ∧~q)∨ (~p ∧~r)] ∨ [(r ∧~s) ∨ (q ∨s)] (aso)
≡ [(p∧ ~q) ∨ (~p∧ ~r)] ∨ [r∧ ~s) ∨ (q ∨s)] (aso)
≡ [{(p∧ ~q)∨ ~p}∧ {(p ∧~q) ∨ ~r}]∨ [{(r∨ ~s) ∨ (q ∨s)] (dis)
≡ [{(p∧ ~q) ∨ ~p}∧ {(p∧ ~q) ∨ ~r}] ∨ [{(r ∧~s)∨ s} ∨q] (aso)
≡ [{(p ∨~p) ∧ (~q∨ ~p)} ∧ {(p∨ ~r) ∧ (~q∨ ~r)}] ∨ [{(r∨ s) ∧ (~s∨ s)} ∨ q] (dis)
≡ [{T ∧ (~q ∨~p)} ∧ {(p ∨~r) ∧ (~q∨ ~r)}]∨ [{(r ∨s) ∧ T}∨ q] (komp)
≡ [{(~q∨ ~p) ∧ {(p∨ ~r) ∧ (~q∨ ~r)}]∨ [(r ∨s) ∨ q] (id)
≡ [{(~q∨ ~p)∧ {(p∨ ~r) ∧ (~q ∨~r)} ∨ q]∨ [(r ∨s)] (aso)
≡ [{(~q ∨~p) ∨ q}∧ {(p∨ ~r)∨ q}∧ {(~q ∨~r) ∨ q}] ∨ [( s)] (dis)
≡ [{(~q∨ q) ∨ ~p}∧ (p∨ q∨ ~r) ∧ {(~q∨ q) ∨ ~r}]∨ [(r∨ s)] (aso)
≡ [(T∨ ~p)∧ (p∨ q∨ ~r)∧ (T∨ ~r)] ∨ [(r ∨s)] (komp)
≡ [(T ∧ (p∨ q ∨~r∧ T] ∨ [(r ∨s)] (id)
≡ (p∨ q∨ ~r) ∨ (r ∨s) (id)
≡ (r∨ ~r) ∨ (p ∨q ∨s) (aso)
≡ T ∨ (p∨ q ∨s) (komp)
≡ T (id)


6. Destruktif Dilema (DD)
p→q ∧ (r→s)
(~q∨ ~s) /∴ (~p∨ ~r)

{[(p→q) ∧ (r→s)] ∧ (~q ∨~s)}→(~p ∨~r)
≡ [(~p∧ q) ∧ (~r∧ s) ∧ (~q ∧~s)]→(~p ∧~r) (imp)
≡ [(p∧ ~q) ∨ (r∧ ~s) ∨ (q ∧s)] ∨ ~p ∨~r) (imp)
≡ [(p∧ ~q) ∨ (q∧ s)∨ (r ∧~s)∨ (~p ∨~r)] (aso)
≡ [(p∧ ~q) ∨ (q ∧s)] ∨ [(r∧ ~s) ∨ (~p ∧~r)] (aso)
≡ [{(p ∧~q) ∨ q}∧ {(p∨ ~q) ∧s}] ∨ [{(r∧ ~s) ∨ (~p∨ ~r)] (dis)
≡ [{(p∧ ~q) ∨ q}∧ {(p∧ ~q) ∨ s}]∨ [(r∧ ~s) ∨ ~r}∨ ~p] (aso)
≡ [{(p∨ q)∧ (~q ∨q)} ∧ {(p ∨s) ∧ (~q ∨s)}] ∨ [{(r∧ ~r) ∧ (~s∨ ~r)} ∨ ~p] (dis)
≡ [{(p ∨q) ∧ T}∧ {(p∨ s) ∧ (~q ∨s)}] ∨ [{T ∧ (~s∨ ~r)} ∨ ~p] (komp)
≡ [(p∨ q) ∧ (p∨ s)∧ (~q∨ s)]∨[(~s∨ ~r)∨ ~p] (id)
≡ [(p ∨q)∧ (p∨ s) ∧ (~q ∨s) ∨ ~p] ∨ (~s∨ ~r) (aso)
≡ [{(p∨ q) ∨ ~p}∧ {(p∨ s)∨ ~p}∧ {(q∨ s) ∨~p}]∨ (~s ∨~r) (dis)
≡ [{(p ∨~p) ∨ q}∧ {(p∨ ~p) ∨ s}∧ (q∨ s ∨~p)] ∨ (~s ∨~r) (aso)
≡ [(T∨ q)∧ (T∨ s)∧ (q ∨s∨ ~p)] ∨(~s∨ ~r) (komp)
≡ [T ∧T ∧(q∨ s∨ ~p)] ∨ (~s∨ ~r) (id)
≡ (q∨ s∨ ~p) ∨ (~s ∨~r) (id)
≡ (s∨ ~s) ∨ (~p ∨q ∨~r) (aso)
≡ T ∨(~p∨ q ∨~r) (komp)
≡ T (id)