jawaban lm 12 novreza
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Tugas BroTRANSCRIPT
Nama : Novreza RidwansyahNIM : 14523243Kelas: E
1. A. (x)A(x) (y)A(y) (x)A(x) (y)A(y) |(y)A(x)(y)A(y) |(y) A(y) | A(a) | A(a)B. (x)A(x) (y)A(y) (x)A(x) (y)A(y) | (x)A(x) (y)A(y) | (y) A(y) | A(a) | A(a) C. (x)(A(x) B(x)) (x)B(x) (x)(A(x) B(x)) (x)B(x) | (x)(A(x) B(x)) (x)B(x) | (x)B(x) | B(a) | (A(a)B(a) | A(a) B(a)
D. (x)((A(x) B(x)) A(x)) (x)((A(x) B(x))A(x) | (x)(A(x) B(x)) (x)A(x) | (x) A(x) | A(a) | A(a) B(a) | A(a) B(a) E. (x)((A(x) B(x)) A(x)) (x)((A(x) B(x))(x)A(x) | (x)(A(x) B(x) (x)A(x) | (x) A(x) | A(a) | A(x) B(a)
A(a) B(a)
F. (x)(A(x) (A(x) B(x)) (x)(A(x) (x)(A(x)B(x)) | (x)(A(x)) (x)(A(x)B(x)) | (x) (A(x)B(x)) | A(a) | (A(a) B(a) | A(a) B(a) G. (x) (A(x) (-A(x) B(x)) (x) (A(x) (x)(-A(x) B(x)) | (x) A(x) (x)(-A(x) B(x)) | (x) ( A(x) B(x)) | A(a) | A(a) B(a)
A(a) B(a)
2. a. Premis 1 : (x) ((B(x) M(x))E(x)) Premis 2 : B(m) M(m) Kesimpulan : E(m)(x)((B(x) M(x))E(x)), B(m) M(m)) E(m) Strategi pembalikan : (x)((B(x) M(x))E(x)), B(m) M(m)) E(m)Jawab : (x)((B(x) M(x))E(x)) B(m) M(m) E(m) | (B(x) M(m))E(x)
(B(m) M(m))E(x)
(B(m) M(m) B. Premis 1 : (x) ((B(x) M(x))E(x)) Premis 2 : B(m) M(m) Kesimpulan : E(m)(x)((B(x) M(x))E(x)), B(m)M(m)) E(m) Strategi pembalikan : (x)((B(x) M(x))E(x)) B(m)M(m)E(m)Jawab : (x)((B(x) M(x))E(x)) B(m) M(m) E(m) | (B(m) M(m))E(m)
(B(m) M(m)) E(x)
(B(m) M(m) C. Premis 1 : (x) ((B(x) m(x))E(x)) Premis 2 : B(m) M(m) Kesimpulan : E(m)(x) ((B(x) m(x))E(x)),(B(m)M(m)) E(m) Strategi pembalikan : (x) ((B(x) m(x))E(x)) (B(m)M(m)) E(m)Jawab : (x) ((B(x) m(x))E(x) B(m) M(m) E(m) | (B(m) M(m))E(m)
(B(m) M(m)) E(m)
(B(m) M(m)
d. Premis = (x) ((B(x) -> ((B(x) |- m(a) -> E(x) ) Premis = B(m) ~m(x) Kesimpulan = E(m)(x) ((B(x) ~m(x) -> E(x) , B(m) ) ~m(x) |- E(m) Strategi pembalikan(x) ((B(x) ~m(x) -> E(x) , B(m) ) ~m(x) ~ E(m)Jawab : (x) ((B(x) ~m(x) -> E(x)B(m)~m(m)~E(m) |(B(m) ~m(m) -> E(m)~(B(m) ~m(m) E(m) ~(B(m) ~m(m) 3. a. (x)((A(x) B(x)) (A(x) C(x)) (x)(B(x) C(x)) (x)((A(x) B(x)) (A(x) C(x)) ~(x)(B(x) C(x)) (x)(A(x) B(x)) (A(x) C(x) )| (x) ~ (B(x) C(x)) |~ (B(a) -> c(a) )| B(a) ~c(a)
~ (A(a) v B(a)) A(a) C(a) | |~ A(a) A(a)~B(a) C(a)3. b.. (x) ((A(x) B(x)) (A(x) B(x)) (c(x) D(x))) (x) (A(x) D(x)) (x) ((A(x) B(x)) (A(x) B(x)) -> (c(x) D(x))) B(x) (A(x) -> D(x) )(x) ((A(x) B(x)) A(x) B(x)) (c(x) D(x))) ~ (x) (A(x) -> (A(x) D(x))(x) ~ (A(x) | A(x) ~ D(x
~ ( ( A(x) (A(x) v B(x) ( C(x) v D(x) | C(x) D(x)
~ (A(x) B(x) ) ~(A(x) V B(x) |~ A(x)~ A(x) ~B(x)~B(x)
c. (x) ((A(x) (B(x) -C(x))) (B(x)) (A(x) C(x)))) (x) (A(x) B(x) -D(x)) (x) (C(x) D(x)) (x) ((A(x) (B(x) ~C(x))) (B(x)) (A(x) C(x)))) (x) (A(x) B(x) ~D(x)) ~ (x) (C(x) D(x) ) |(x) ~ (C(X) D(x) ) |A (a) B (a) ~D(a) |A (a) -> ((B(a) ~ c(a) )B(a) -> (A(a) -> c(a))~ C(a) ~ D(a)
~ A(a) B(a) ~ c(a) ) ~ B(a) A(a) -> C(a) | B(a) ~A(a) C(a) ~C(a)