Membuktikan A ∩ B=B ∩ A

Show that:1. A ∩ B=B ∩ A2. (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 ∩...

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Modus Ponen (MP)

1. Modus Ponen (MP)p⇒qp∴qPembuktian:[(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 :Argumenp⇒qp∴q Argumen sah2. Modus Tollens (MT)p⇒q~q∴~pPembuktian ;[(p⇒q)∧~q]⇒~p⇔ ~[(~p∨q)∧~q]∨~p (Imp)⇔ [(p∧~q)∨q]∨~p (DM)⇔ [(p∨q)∧(~q∨q)]∨~p (Dist)⇔ [(p∨q)∧T]∨~p...

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