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Ujian Take Home Exam Universitas Terbuka Logika Informatika 2022.2 sebagai referensi belajar
Typology: Essays (university)
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UJIAN THE (Take Home Exam)
F 2 : if true then ( if not Q then ( P or S ) else not ( P and S )) else ( P and Q )
bernilai false.
Tentukan interpretasi yang membuat F 2 bernilai false!
Jawaban
Diketahui
2
: if true then ( if not Q then ( P or S ) else not ( P and S )) else ( P and Q )
bernilai false dengan aturan If-then-else ketika
(i) If true bernilai true
(ii) if not Q then ( P or S ) else not ( P and S ) bernilai false
(iii) ( P and Q ) bernilai true
(i) If true bernilai true
(ii) Aturan If-then-else
Pada if not Q then ( P or S ) else not ( P and S ) bernilai false ketika
Not Q bernilai false, ( P or S ) bernilai true, not ( P and S ) bernilai false
a. Dari aturan Not Q bernilai false,
Q bernilai true
b. Dari aturan P or S bernilai true jika
b.1) P bernilai false, S bernilai true
b.2) P bernilai true , S bernilai true
b.3) P bernilai true, S bernilai false
c. Dari aturan not (P and S) bernilai false
Ketika P and S bernilai true jika
c.1) P bernilai true , S bernilai true
(iii) Aturan and pada (P and Q) bernilai true, jika
P bernilai true , Q bernilai true
Jadi interpretasi yang membuat F 2
bernilai false adalah {Ptrue, Q true, S true}
( P or ( Q and S ))
Jawaban
Perlihatkan dengan metode Proof by Falsification (PbF) bahwa kalimat E berikut valid. E: if P then
( P or ( Q and S ))
Keterangan:
P bernilai Benar, ( P or ( Q and S )) bernilai Salah
P bernilai salah, (Q and S) bernilai salah
Ditemukan kontradiksi pada nilai P sehingga kalimat E adalah valid
{( P and not R ) or (if Q then ( P or ( Q and R )))} ◁ {( P and R ) Q , ( Q and R ) ( P and S ), ( not R )
( P or S )}.
Jawaban
{(P and not R) or (if Q then (P or (Q and R)))} ◁ {(P and R) Q, (Q and R) (P and S), (not R)
(P or S)}.
Identifikasi kalimat-kalimat
F: {(P and not R) or (if Q then (P or (Q and R)))} ◁ {(P and R) Q, (Q and R) (P and S), (not R) (P or
S)}.
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2
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f i
(d 1
, d 2
) = d 1
d 2
fj (d 1 , d 2 ) = d 1 d 2 = 9 x 2= 18