A A A B A B A B A B A B A B A B A B A B A B 1 A B A B A B 1.3 (1.3) A B B A 1 2

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1 Morality mod Science (1.1) 1 2 A 1, A 2,..., A n B A 1, A 2,..., A n B A 1, A 2,..., A n B A 1, A 2,..., A n B (1.2) 1 A B 2 B A 1.2 A B minao.kukita@gmail.com 1

2 A A A B A B A B A B A B A B A B A B A B A B *1 A B A B A B 1.3 (1.3) A B B A *1 2

3 (1.4) 1 X Z 2 Y Z X Y X Y Z Z (1.5) 1 2 (1.6) (1.7) 1 2 (1.8) Exercise

4 (1.9) A B C C A B A B P Q P Q P Q (1.10) P Q Q P A B (1.11) A A 1.5 4

5 1. p, q, r, p 1, q 1, r 1, A, B A, (A B), (A B), (A B) 3. Form Atom ( p 3 r), (r 2 (q 5 p)) p, (q ), (q 1 r p q r (p (q r)) ((p q) r) x + y z x + (y z),, p q r p (p ((q r) p)) p q r (p (q r)) Exercise p q r q 2. q r p q r p 3. p q r q p r q r Exercise (((p q) r) ((p r) (q r))) 2. (( p q) (( p q) p)) 3. (((p q) p) p) 4. ( (p q) ( p q)) 5. ((p q) ((r p) (r q))) 6. ((p (q r)) ((p q) (p r))) p, q 1, q 5 r 2 5

6 1 0 Definition 2.1 ( ). f f A, B 1. f( A) = 1 f(a) = 0 2. f(a B) = 1 f(a) = 1 f(b) = 1 3. f(a B) = 1 f(a) = 1 f(b) = 1 4. f(a B) = 1 f(a) = 0 f(b) = 1 v, v,... S v S S 1 0 v {P1,P 2,...,P n} P 1, P 2,..., P n 1 v P v[p ] P 1 v v[ P ] P 0 v Exercise 2.2. P v(p ) = v (P ) v, v Definition 2.3 ( ). A 1, A 2,..., A n, B v v(a 1 ) = v(a 2 ) = = v(a n ) = 1 v(b) = 1 B A 1, A 2,..., A n A 1, A 2,..., A n = B v v(b) = 1 B = B v v(a) = v(b) A B A B Example 2.4. Γ, = 1. Γ, A, B, = C = Γ, B, A, = C 2. Γ = A = Γ, B = A 3. Γ, A, A = B = Γ, A = B 6

7 4. = A A 5. A B, A B = A 6. A, B = A B 7. A 1 A 2 = A i (i = 1, 2) 8. A i = A 1 A 2 (i = 1, 2) 9. A B, A C, B C = C 10. A = B = A B 11. A B, A = B 12. A = A 13. = A A 14. = (A A) 15. = A (B A) 16. = A (A B) 17. = (A B) (B A) 18. = (A B) (B C) 19. = ((A B) A) A Exercise material implication A B A B

8 A B A B A B A B A B (1)-(4) A B A B (1) (2) (3) (4) A B A B (2) (1)(4) (1)(4) (3) (3) (3) A B A B B A A B (A B) (B A) A B B A A B A B 8

9 formal implication x 4 x 2 x x x x strict implication 9

10 3.1 A A 1. p, q, r, p 1, q 1, r 1, A, B A, (A B), (A B), (A B), A 3. A A A A A A Form 3.2 (1) (2) (3) (4) 10

11 Definition 3.1 ( ). W w W v w : Form {0, 1} A, B 1. v w ( A) = 1 v w (A) = 0 2. v w (A B) = 1 v w (A) = 1 v w (B) = 1 3. v w (A B) = 1 v w (A) = 1 v w (B) = 1 4. v w (A B) = 1 v w (A) = 0 v w (B) = 1 5. v w ( A) = 1 w W v w (A) = 1 6. v w ( A) = 1 w W v w (A) = 1 v = {v w : w W } v) S = v) w W v w (A) = 1 A S w S, w = A w = S A w = A Definition 3.2 ( ). A S = v) = A S = A = S A A 1, A 2,..., A n, B S = S A i A i (1 i n) = S B B A 1, A 2,..., A n A 1, A 2,..., A n = B Theorem 3.3 ( ). A, B 1. A = A 2. = A = A 3. (A B) = A B 4. (A B) = A B 5. (A B) = A B 6. A = A 7. A = A 8. (A B) = A B 9. A B = (A B) 10. A = A 11. A = A 11

12 12. A = A 13. A = A 14. A = A Proof 2 = A S = v) S = A = A w W S, w = A S (W, w, v) w S S, w = A S = A = A 3 S = v) S = (A B) S = A w W w = A B w = A w = B w W w = B S = B S = A B Exercise A B A B (A B) A B A B Theorem 3.5 ( ). A, B, C 1. = A (B A) 2. = A (A B) 3. A = B = A B 4. = A A 5. A B, A = B 6. A B, B C = A C 7. A B B A 8. A B, A C = A (B C) 9. A C, B C = (A B) C 10. (A B) C (A C) (B C) 11. = A (B A) 12. = A (A B) 13. = (A B) (B A) 14. = (A B) (B C) 15. = ((A B) A) A 12

13 16. A B C = A (B C) 17. A (B C) = B (A C) 18. = A (B A B)) Proof 1 1 S = v) = A (B A) = ( A (B A) w W S, w = A (B A) S, w = A S, w = (B A) w W S, w = B A S, w = B S, w = A S, w = A S, w = A 11 W = {@, w}, (p) = 1, v w (p) = 0, v w (q) = 1 S = v) S, w = p S, w = q S, w = q p = (q p) = p = p (q p) = (p (q p)) = p (q p) S Exercise = A B C A B C 17 A B C B A C 18 A B A B 13

14 3.4 M M A A A A Definition 3.7 ( ). W R W W R W w W v w : Form {0, 1} A, B 1. v w ( A) = 1 v w (A) = 0 2. v w (A B) = 1 v w (A) = 1 v w (B) = 1 3. v w (A B) = 1 v w (A) = 1 v w (B) = 1 4. v w (A B) = 1 v w (A) = 0 v w (B) = 1 5. v w ( A) = 1 wrw w W v w (A) = 1 6. v w ( A) = 1 wrw w W v w (A) = 1 v = {v w : w W } R, v) A A = A A = A

15 Definition 3.8 ( ). W R W W R reflexive w W wrw R symmetrical w, w W wrw w Rw R serial w W w W wrw R transitive w, w, w W wrw w Rw wrw R assymmetrical w, w W wrw w Rw w = w R R R R R Euclidean w, w, w W wrw, wrw w Rw R equivalence R Theorem 3.9. S = (W, v) 1. R = A S = A A 2. R = A S = A A 3. R = A S = A A 4. R = A S = A A 5. R = A S = A A Proof 1 S = (W, v) = = A Exercise Definition K KD KT KTB S4 S5 Theorem K KD 15

16 2. K KT 3. KT KTB 4. KT S4 5. KTB S5 6. S4 S5 7. S5 M Proof 1 K KD K K KD A A KD K S = = A = = A A S5 S = (W, v) S = v ) W = {w : w w W p Atom v w(p) = v w (p) A w W S, w = A S, w = A (1) A A B B = S = w W S, w = B w W S, w = B = B = 1 S5 S M S M S = v) S5 S (W, W v) A w W S, w = A S, w = A (2) A Exercise Exercise A n A n A A n n 0 2. A A. 16

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01_教職員.indd T. A. H. A. K. A. R. I. K. O. S. O. Y. O. M. K. Y. K. G. K. R. S. A. S. M. S. R. S. M. S. I. S. T. S. K.T. R. T. R. T. S. T. S. T. A. T. A. D. T. N. N. N. Y. N. S. N. S. H. R. H. W. H. T. H. K. M. K. M.