Membuktikan A ∩ B=B ∩ A

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
⇔ x ∈ (B ∩ A)
so A ∩ B ⊂ B ∩ A...........( I )

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.............( II )
From (!) and (II) we conclude that A ∩ B ⊂ B ∩ A

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Modus Ponen (MP)

1. Modus Ponen (MP)

p⇒q
p
∴q

Pembuktian:
[(p⇒q)∧p]⇒q
⇔ ~[(~p∨q)∧p]∨q (Imp)
⇔ [(p∧~q)∨~p]∨q (Komp.DM)
⇔ [(p∨~p)∧(~p∨~q)]∨q (Dist)
⇔ [T∧(~p∨~q)]∨q (Komp)
⇔ (~p∨~q)]∨q (Id)
⇔ ~p∨(~q∨q) (As)
⇔ ~p∨T (Komp)
⇔ T (Id)

Kesimpulan :
Argumen
p⇒q
p
∴q
Argumen sah

2. Modus Tollens (MT)

p⇒q
~q
∴~p

Pembuktian ;
[(p⇒q)∧~q]⇒~p
⇔ ~[(~p∨q)∧~q]∨~p (Imp)
⇔ [(p∧~q)∨q]∨~p (DM)
⇔ [(p∨q)∧(~q∨q)]∨~p (Dist)
⇔ [(p∨q)∧T]∨~p (Komp)
⇔ (p∨q)∨~p (Id)
⇔ (p∨~p)∨(q∨~p) (Dist)
⇔ T∨(q∨~p) (Komp)
⇔ T (Id)

Kesimpulan :
Argumen
p⇒q
~q
∴~p
Argumen sah

3. Silogisme

p⇒q
q⇒r
∴p⇒r

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