TUGAS 5 PDM

Show that A ⊂ B ⇔ A ∩ B = A
Proof:
(i) Show that A ⊂ B ⇒ A ∩ B = A
We have A ⊂ B
it means that ∀ 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 ∀ 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 ∈ A
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

Versi kontradiksi:
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 ∈ (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 ⊂ B
We 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 ⊂ B

Final conclusion : A ⊂ B ⇔ A ∩ B = A

TUGAS 4 PDM

Show that:

1. A ∩ B=B ∩ A
2. (A ∩ B) ∩ C=A ∩ (B ∩ C)

Answer:
1. Proof:
Show that A ∩ B ⊂ B ∩ A
take any x ∈ (A ∩ B)
obvious x ∈ (A ∩ B)
≡ x ∈ A ∧ x ∈ B
≡ x ∈ B ∧ x ∈ A (komutatif)
≡ x ∈ (B ∩ A)
so A ∩ B ⊂ B ∩ A.........................( 1 )

Show that B ∩ A ⊂ A ∩ B
take any x ∈ (B ∩ A)
obvious x ∈ (B ∩ A)
≡ x ∈ B ∧ x ∈ A
≡ x ∈ A ∧ x ∈ B (komutatif)
≡ x ∈ (A ∩ B)
so B ∩ A ⊂ A ∩ B..........................( 2 )
From (1) and (2) we conclude that A ∩ B ⊂ B ∩ A


2. Proof :
Show that (A ∩ B) ∩ C ⊂ A ∩ (B ∩ C)
take any x ∈ [(A ∩ B) ∩ C]
obvious x ∈ [(A ∩ B) ∩ C]
≡ x ∈ (A ∩ B) ∧ x ∈ C
≡ (x ∈ A ∧ x ∈ B) ∧ x ∈ C
≡ x ∈ A ∧ (x ∈ B ∧ x ∈ C) (assosiatif)
≡ x ∈ A ∧ x ∈ (B ∩ C)
≡ A ∩ (B ∩ C)
so (A ∩ B) ∩ C ⊂ A ∩ (B ∩ C).....................(1)

Show that A ∩ (B ∩ C) ⊂ (A ∩ B) ∩ C
take any x ∈ [A ∩ (B ∩ C)]
obvious x ∈ [A ∩ (B ∩ C)]
≡ x ∈ A ∧ x ∈ (B ∩ C)
≡ x ∈ A ∧ (x ∈ B ∧ x ∈ C)
≡ (x ∈ A ∧ x ∈ B) ∧ x ∈ C (assosiatif)
≡ x ∈ (A ∩ B) ∧ x ∈ C
≡ (A ∩ B) ∩ C
so A ∩ (B ∩ C) ⊂ (A ∩ B) ∩ C...................... (2)
From (1) and (2) we conclude that (A ∩ B) ∩ C ⊂ A ∩ (B ∩ C)

TUGAS 3 PDM

1. Modus Ponen
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
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)] (kom)
≡ (p⇒ q) ⇒ [(q ∧ ¬r) ∨ r] ∨ ¬p (as0)
≡ (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
(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 Delema
(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)] (dist)
≡ [{(p ∧ ¬q) ∨ ¬p} ∧ {(p ∧ ¬q) ∨ ¬r}] ∨ [{(r ∧ ¬s) ∨ s} ∨ q] (aso)
≡ [{(p ∨ ¬p) ∧ (¬q ∨ ¬p)} ∧ {(p ∨ ¬r) ∧ (¬q ∨ ¬r)}] ∨ [{(r ∨ s) ∧ (¬s ∨ s)} ∨ q] (dist)
≡ [{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}] ∨ (r ∨ s) (dist)
≡ [{(¬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. Distruktif Delema
(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)] (dist)
≡ [{(p ∧ ¬q) ∨ q} ∧ {(p ∧ ¬q) ∨ s}] ∨ [{(r ∧ ¬s) ∨ ¬r} ∨ ¬p] (aso)
≡ [{(p ∨ q) ∧ (¬q ∨ q)} ∧ {(p ∨ s) ∧ (¬q ∨ s)}] ∨ [{(r ∨ ¬r) ∧ (¬s ∨ ¬r)} ∨ ¬p] (dist)
≡ [{(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) (dist)
≡ [{(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)