Soal No 6

6.    Show that (i)   ̴x eq   ̴(xvy) and (ii) x eq (x→y) →(x→y) from these find the equivalent formulae of x˄y, x→y and x↔y as iterated compositions of joint negations

Answer : (i)    ̴ x eq x→x

 

Download : Jawaban Soal no 6

Soal no 5

5.

Corresponding to the statement 'neither X nor Y', the joint negation

X i Y is defined by the truth table.

Show that (i) X i Y eq ~ (X v Y) and (ii) X eq (X I X) I (X I X).

download : jawaban soal no 5

Contoh Soal

Buktikan [(p → q) ˄p] → q ek T

Bukti :

Jelas :    [( p →  q ) ˄ p ] → q

                ≡ [( ~p  v  q) ˄ p ] → q                       (hk. implikasi)

                ≡~ [( ~p  v  q ) ˄ p ] v q                      (hk. implikasi)

                ≡ [~(~p  v  q ) v ~p ] v q                     (hk. DM)

                ≡ [(~(~p)  ^  ~q ) v ~p ] v q                (hk. DM)

                ≡ [(p  ^  ~q ) v ~p ] v q                       (hk. Komplemen)

                ≡ [(p  v  ~p ) ^(~q v  ~p) ] v q            (hk. Distributis )

                ≡ [T ^(~q v  ~p) ] v q                          (hk. Komplemen)

                ≡ (~q v  ~p)  v q                                  (hk. Identitas)

                ≡ ~q v ( ~p  v q)                                  (hk. Asosiatif)

                ≡ ~q v ( q  v ~p)                                  (hk. Komutatif)

                ≡(~q v q) v  ~p                                    (hk.Asosiatif)

                ≡ T v ~p                                               (hk. Komplemen)

                ≡ T                                                       (hk. Identitas)

Jadi [(p → q) ˄p] → q ek T BENAR! yeeeeeeeeeeeee
by Kelompok 6

File : Pembuktian Contoh Soal ( kel 6 )

Mind Map Statment Calculus

Peta konsep diatas , dibuat berdasarkan hasil dari presentasikan kel 1yaitu materi Logika Matematika.

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