Kelompok 7
1 1. Anninditya Atika K.D
2. Devian Putri
3. Sara Nurul Hidayah 4101412136
3. Sara Nurul Hidayah 4101412136
SOAL :
1. [(p ʌ q) ʌ p] → q ek T
2. [(p → q) ʌ p] → q ek T
3. [(p → q) ʌ ~ q] → ~ p ek T
JAWAB :
1.
Bukti :
Jelas : [(p ʌ q) ʌ p] → q


(DM) Ξ [(~ p v ~ q) v ~ p] v q
(assosiatif) Ξ [(~ p v (~ q v ~ p)] v q
(komutatif) Ξ [(~ p v (~ p v ~ q)] v q
(assosiatif) Ξ [(~ p v ~ p) v ~ q)] v q
(idempoten) Ξ ~ p v (~ q v q)
(komplemen) Ξ ~ p v T
(identitas) Ξ T (terbukti)
2.Bukti :
Jelas : [( p → q ) ʌ p ] → q


(DM) Ξ [(p ʌ ~ q) v ~ p] v q
(assositif) Ξ [p ʌ (~ q v ~ p)] v q
(komutatif) Ξ [p ʌ (~ p v ~ q)] v q
(assosiatif) Ξ [(p ʌ ~ p) v ~ q)] v q
(komutatif) Ξ [F v ~ q] v q
(assosiatif) Ξ F
v (~ q v q)
(komplemen) Ξ F v T
(identitas) Ξ T (terbukti)
3.Bukti :

(implikasi) Ξ [(p → q) ʌ ~ q] v ~ p

(DM) Ξ [(p ʌ ~ q) v q] v ~ p
(assosiatif) Ξ [p ʌ (~ q v q)] v ~ p
(komplemen) Ξ [p ʌ T] v ~ p
(identitas) Ξ p v ~ p
(komplemen)
Ξ T
(terbukti)
SOAL :
1.Corresponding
to the statement neither X nor Y , the joint negation X↓Y is defined by the truth table
X
|
Y
|
X↓Y
|
T
|
T
|
F
|
F
|
T
|
F
|
T
|
F
|
F
|
F
|
F
|
T
|
Show
that (i) X↓Y eq ~(X v Y) and (ii) X eq (X↓X)
↓ (X↓X)
2.Show
that (i) ~X eq X↓X and (ii) X v Y eq (X↓Y)→ (X↓Y),from
these result find the equivalent formulae of X ʌ Y, X → Y and X ↔ Y as iterated
compositions of joint negations.
JAWAB:
1.
(i) ) X↓Y eq ~(X v Y)
X
|
Y
|
X v Y
|
X ↓ Y
|
~ (X v Y)
|
T
|
T
|
T
|
F
|
F
|
F
|
T
|
T
|
F
|
F
|
T
|
F
|
T
|
F
|
F
|
F
|
F
|
F
|
T
|
T
|
(ii)
X eq (X↓X) ↓ (X↓X)
X
|
X↓X
|
(X↓X) ↓ (X↓X)
|
T
|
F
|
T
|
F
|
T
|
F
|
T
|
F
|
T
|
F
|
T
|
F
|
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