Upcoming SlideShare
×

# ตรรกศาสตร์เบื้องต้น

1,210 views

Published on

Published in: Spiritual, Technology
1 Comment
1 Like
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
Your message goes here
• ขอบคุณค่ะ

Are you sure you want to  Yes  No
Your message goes here
Views
Total views
1,210
On SlideShare
0
From Embeds
0
Number of Embeds
3
Actions
Shares
0
85
1
Likes
1
Embeds 0
No embeds

No notes for slide

### ตรรกศาสตร์เบื้องต้น

1. 1. F 31201 ก 4 WiLa 1 1 ก F F1.1 F (Propositions or Statements) F F ก F F F F F F กF F (truth value) F F ก F T F F F ก F ก F F F ก ก F a, b, c, , z F F 1. ก F ( ) 2. -5 ( ) 3. 0 F ( ) F F F F 1 F กF F F ก F F F F F F 2 ก F F F F ก F F F ก F FF F F 2 ก กF F x+2=5 F x F F F F 1 F F F F1. กF F F ......................................2. 5+ 6 = 12 ..................................3. ก 49 7 -7 ......................................4. x+y+5 =0 ...................................5. F F .....................................6. a+0 = a .................................
2. 2. F 31201 ก 4 WiLa 2 ก 1 F F F F1. 0 F2. F F3. A ∩ B = B ∩4. F5. 1+5 = 86. x + 7 = 87. ก8. ก F F9. F F Fก 410. ก F ก ก ก F11. F12. ก F F ก13. {1,2,3} = {2,3,4}14. π ก15. {0} F16. x x>517. F ก F18. ก ก 419. x y F x+y=y+x20. F
3. 3. F 31201 ก 4 WiLa 31.2 ก F F F กก F F F ... F ... ก F ก F F F กF (connectives) F 0 F F 2 4 F F 3 F 321. ก F F F 5+ 2 = 2 + 5 3 x 1 = 1x 3 F F F F F 5+ 2 = 2 + 5 3 x 1 = 1x 3 p q F p∧q F (truth table) p∧q F q p∧q P T T T T F F F T F F F F 2. ก F F F 1+8=8+1 5( 3 + 7 ) = ( 5 x 3 ) + ( 5 x 7 ) F F F F F 1+8=8+1 5( 3 + 7 ) = ( 5 x 3 ) + ( 5 x 7 ) p q F p∨q F (truth table) p∨q F P q p∨q T T T T F T F T T F F F
4. 4. F 31201 ก 4 WiLa 43. ก F F F ... F ... F 5<7 5+(-3)<7+(-3) F F F ... F ... F F F F 5<7 F 5+(-3)<7+(-3) F p F q F p→q F (truth table) p→q F P q p→q T T T T F F F T T F F T4. ก F F ก F F 5( 7 + 3 ) = 5 x 10 7 + 3 = 10 F F ก F F F F 5( 7 + 3 ) = 5 x 10 ก F 7 + 3 = 10 pก F q F p↔q F (truth table) p↔q F P q p↔q T T T T F F F T F F F T 1. F ก F F กF F F (atomic statement) 2. ก F ก กก FF F F F ก5. F F 2+3=5 2+3≠5 F 2<3 2<3(F F 2 F F กF 3 F 2 Fก 3 กก F 3 Fก F 2≥3 )
5. 5. F 31201 ก 4 WiLa 5 Fp F ~p F ~p F P ~p T F F T ก 21. F F F ก F 1) 4 5 F ..................................................................................... 2) 3 Fก 4 3 F ก F 4 ..................................................................................... 3) 4 F F 43 F ..................................................................................... 4) F F F ก ........................................................................ 5) 3 × 5 = 15 ก F 15 ÷ 3 = 5 .....................................................................................2. ก F p F 3 q F 3 r F 2 F s F 2 ก FF 1) p ∧ ∼q 2) r⇒s 3) ∼r ⇔ s 4) (p ∧ q) ⇒ r 5) q ⇔ (r ∨ ∼ p) ..3. ก F FF 1) 4 5 F 2) 2 Fก 3 2 กก F 3 ..... 3) F 7 F F 72 4) 2<5 ก F 3 > 5 ....... 5) F {1 , 2} = {2 , 1} F {1 , 2} ⊂ {2 , 1} ......
6. 6. F 31201 ก 4 WiLa 61.3 ก F F F F F ก F F F F F ก F F F F F 1 F F F Fp F Fq ก F F p∧q กp q F p∧q F F F 2 ก F a, b c F F F (a ∧ b) ∨ c กa b F a∧b ก a∧b c F (a ∧ b) ∨ c ( a ∧ b) ∨ c T T F T T ก F F F F F F 3 F ~ ( a → ~ b) a,b F F กb F ~b กa ~b F a →~ b ~ ( a → ~ b) F ~ (a → ~ b ) T T F F T
7. 7. F 31201 ก 4 WiLa 7 F 4 ก Fp q r s F [ ( p ∧ q) ∨ r ] → ( p ∧ s) [ ( p ∧ q ) ∨ r ] → ( p ∨ s ) T F F F T T F T T F [ ( p ∧ q) ∨ r ] → ( p ∧ s) F ก 31. ก F P F F Q F F R F F S F F F FF 1. [ P ∧ (~ Q)] ↔ ( P ∨ S) 2. ( P → Q) → (S ∨ R) 3. [ P ∨ (~ R)] → S 4. [( P ∨ Q) ∧ (~ R)] → Q 5. ( P ∧ Q) ∨ (~ R) 6. ( P ↔ R) → (Q ∨ S) 7. Q ↔ [( P ∧ S) ∨ R ] 8. ~ ( P ∧ Q) ↔ [(~ P) ∨ (~ Q)]2. ก F P,Q,R S F P Q F R ∧ (~ S) F F F F 1. ( P ∨ Q) → ( P∧ ~ R) 2. [( P ∧ S) ∨ (~ R)] → ( R ∧ Q) 3. ( P ↔ R ) → (Q → S) 4. [( R ∧ Q) ∨ (S → P)] → [( P ∧ (S ∨ Q)]3. F P,Q,R,S F F [( P → Q)].V ( R∨ ~ S ) F P,Q,R,S4. F P,Q,R,S F P∨Q F (S ∨ R ) ∨ Q F F P,Q,R,S
8. 8. F 31201 ก 4 WiLa 81.4 ก F F ก F F F ก F F F F F F F F ก F F F F 1 F F F ( p → q ) ∧ ∼q p q ∼q p→q ( p → q ) ∧ ∼q T T F T F T F T F F F T F T F F F T T T F 2 F F F ( p ∧ ∼q ) ∨ ∼r p q r ∼q ∼r p ∧ ∼q ( p ∧ ∼q ) ∨ ∼r T T T F F F F T T F F T F T T F T T F T T T F F T T T T F T T F F F F F T F F T F T F F T T F F F F F F T T F T F F F F F 3 F F F F ( p ∧ ∼q ) ∨ ∼r (p → q) ∧ ∼q T T T F F T F F F T F T T F F F T F T T
9. 9. F 31201 ก 4 WiLa 9 F 4 F F F F ( p ∧ ∼q ) ∨ ∼r (p ∧ ∼q ) ∨ ∼r T F F F F T F F T T T T T T F T T T T T F F F F F F F F T T F F T F F F F T T T1.5 F ก ก F F F F ก ก Fก F F ก F ก F กF F F ก F p→q ก ~ p∨q ก ก F F F 1 ก Fp q F F p→q ~p∨q p q p→q ~p ~ p∨q T T T F T T F F F F F T T T T F F T T TQ F p→q ~p∨q F ก กก F Fp,q∴ F p→q ก F ~p∨q ก F F Fa ก Fb F F a≡b
10. 10. F 31201 ก 4 WiLa 10F 2 F p ∧∼q ก ∼( q→p ) ก F p q ∼p ∼p ∧q q→p ∼( q→p ) → T T F F T F ก T F F F T F p ∧∼q ก ∼( q→p ) F F T T T F T ก กก F F T F T F ∴ p ∧∼q ≡ ∼( q→p ) F ก Fก ก F p, q r F 1. p ∧ p ≡ p 2. p ∧ q ≡ q ∧ p 3. p ∨ q ≡ q ∨ p 4. p ∧ q ≡ q ∧ p 5. p → q ≡ ∼( p ∨ q ) ≡ ∼ q → ∼p 6. p ↔ q ≡ ( p → q ) ∧ (q → p ) ≡ ∼ p ↔ ∼q 7. ∼(∼ p ) ≡ p 8. ∼ (p ∧ q) ≡ ∼p ∨ ∼ q 9. ∼(p ∨ q) ≡ ∼p ∧ ∼ q 10. ∼ ( p → q ) ≡ p ∧ ∼q 11. ∼ ( p ↔ q ) ≡ ∼p ↔ q ≡ p ↔ ∼q 12. p∧ (q ∨ r ) ≡ ( p∧q ) ∨ ( p∧r ) p∨ (q∧r ) ≡ ( p∨q ) ∧ ( p∨r ) 13. p→ ( q ∧ r ) ≡ ( p→q ) ∧ ( p→r ) p→ ( q ∨ r ) ≡ ( p→q ) ∨ ( p→r ) 14. p→ ( q → r ) ≡ ( p∧q ) → r 15. ( p→q ) ∧ ( q→r ) ≡ p→ r
11. 11. F 31201 ก 4 WiLa 11 ก 41. F F F F 1.1 (p → q) → (∼ p ∧∼ q ) 1.2 ( p ∧∼ q ) ↔ ( q ∨ p ) 1.3 (p∧q) → ( p∨r)2. F F F ก F 2.1 ∼ p ∧ q ก ∼ (q → p) 2.2 p → q ก ∼ p →∼ q 2.3 (p∧q) → r ก p→ ( q → r)1.6 F (TAUTOLOGY) ก F F กก F Fก F F ก F F F F F F 1 ก Fp q F F ( p → q) ∧ p → q p q p→ q ( p → q) ∧ p ( p → q) ∧ p → q T T T T T T F F F T F T T F T F F T F TQ ( p → q) ∧ p → q F กก F Fp q ∴ ( p → q) ∧ p → q Fก F F ก F F F F ก F F F F ก Fก F ก F F FF ก F Fก F F F F FF F F Fก F F F F F
12. 12. F 31201 ก 4 WiLa 12 F 2 F [( p → q) ∧ p] →~ q F F F [( p → q) ∧ p] →~ q F [ ( p → q ) ∧ p ] → ~ q F T F T T T T ก F F [( p → q) ∧ p] →~ q F F FF F Fก ก p q F [( p → q) ∧ p] →~ q F [( p → q) ∧ p] →~ q F F F 3 F ( p ∧ q) → (q ∨ p ) F F F ( p ∧ q) → (q ∨ p ) F ( p ∧ q ) → ( q ∨ p ) F T F T T F F Fก ก F F p q F ก ก Fก FF [( p → q ) ∧ p ] →~ q F [( p → q) ∧ p] →~ q F
13. 13. F 31201 ก 4 WiLa 13 (2) F F A∨B ก A∨B F ก A F B F ก FA∨B F F F F F (3) F F A↔B ก F A↔B F F F F F ก ก Fก F A ก Bก F F A B ก กก ก F A↔B T กก ก F A↔B F1.7 ก F ก F ก F ก F( ) F F ก F F F F F ก F F F ก F F F F F F F Fก F F ก F3 1. ก Fก ก F ก ก F 1. ก ก (modus ponens) 2. ก ก F (modus tollens) p →q p →q p ~q ∴q ∴~ p 3. ก ก (law of syllogism) 4. ก ก ก (disjunctive syllogism) p →q p ∨q q →r ~ p ∴p → r ∴q5. ก ก F (conjunctive inference) 6. ก ก ก (inference by cases) p p →r q q →r ∴p ∧q ∴p ∨q → r7. ก ก F F (law of simplification) 8. ก F F p ∧q p →q ∴p ∴~ q →~ p
14. 14. F 31201 ก 4 WiLa 14 2. ก Fก F ก F ก F F F p1 , p2 ,..., pn F F C F Fก F ก F F F ก F F กF F p1 , p2 ,..., pn F F กF F Cก F F ก F F F ∧ F F ก F → F ก ( p1 , p2 ,..., pn ) → C F ( p1 , p2 ,..., pn ) → C F ก F F ก F (valid) FF ( p1 , p2 ,..., pn ) → C F Fก F FF ก F F (invalid) ก F ก ก F F 1 Fก F F F : 1. p→q 2. ~ q : ~ p ก F F [( p → q )∧ ~ q ] →~ p F p q ~p ~q p→q ( p → q )∧ ~ q [( p → q )∧ ~ q ] →~ p T T F F T F T T F F T F F T F T T F T F T F F T T T T T ก F F กก Fก F F ก F F 2 : 1. F กF ก 2. F F กF :ก F F ก F ก F FF ก F p กF q
15. 15. F 31201 ก 4 WiLa 15 : 1. p → q 2. ~ p : q ก F [( p →q)∧ ~ p] → ( q) F( ) T T= ~ T = ( p → q) p F= q F= p F= q F= p F F F ก p q FF F ก 5 ก F F F1. 1. p → q 2. q → r 3. ∼r ∼p∨r2. 1. F 7 F F 7 F 2 2. 7 F F F 23. 1. F กก ก F 2. F F กก ก F F 3. F F
16. 16. F 31201 ก 4 WiLa 161.8 ก F F F F F F ก ก F F F F x F กก F 3 a+2= 1NOTE F P(x) Q(x) ก F 1 F F 1. ก 2. กก 3. x - 5 = 10 4. F x+2=3 F x-2 = 0 5. π ก 6. 3x = 15 x=3 F F 7. x + 10 8. a + a = 2a F1 ก F F F2 กก F F F3 x - 5 = 10 x F x ∈R F F4 F x+2=3 F x-2=0 x F x ∈R F F5 π ก F F6 3x = 15 x F x ∈R F F7 x + 10 F x F F x ∈R F F F F F8 a + a = 2a F F F
17. 17. F 31201 ก 4 WiLa 171.9 F F 2 1. F ... ก F ก ก ก F x ก x x F F ... ก F ก F ∀ F F P(x) x+2 = 3 ∀ x P (x) ∀ x[x+2 =3 ] x ก x+2 = 3 2. F ... F ก F x x x F F F ... F ก F ∃ F F P(x) x-3 = 5 ∃ xP(x) ∃ x[x - 3 = 5] x x-3 = 5 F ก F ∀x x ก F ก F ∃x x F ก FU ก F F ก FR F ก FQ ก F ก FI Z F ก FN F F F F P(x) x (1) F ∀x[ p(x)] F ก F x p(x) F ก F U F F F ∀x[ p(x)] ก F ก x∈U , P(x) (2) F ∀x[ p(x)] F ก F ก U F F x p(x) F F F
18. 18. F 31201 ก 4 WiLa 18 ∀x[ p(x)] ก F x∈U P(x) (3) F ∃x [p (x ) ] F ก F ก U F F x p(x) F F F ∃x[ p( x)] ก F x∈U P(x) (4) F ∃x [p (x ) ] F ก F x p(x) F ก F UF F F ∃x[ p( x)] ก F ก x∈U , P(x) F 1 F F F 1. ∀ x [ x + 8 ≥ 8 ] U = { 0, 2, 4 } F x∈U ก F x+8≥ 8 x=0 ; 0+8≥8 (T) x=2 ; 2+8≥8 (T) x=4 ; 4+8≥8 (T) F ก F x∈U F x+8≥ 8 ∀ x [ x + 8 ≥ 8 ] ; U = { 0, 2, 4 } F 2. ∀x [ x + 8 > 8 ] U = { 0, 2, 4 } F x∈U ก F x+8>8 x=0 ; 0+8>8 (F) x=2 ; 2+8>8 (T) x=4 ; 4+8>8 (T) F F x∈U F x+8> 8 ∀ x [ x + 8 > 8 ] ; U = { 0, 2, 4 } F 3. ∃x[ x 2 = 2 x] U = { -1, 0, 1 } F x∈U ก F x2 = 2x x = -1 ; (-1)2 = 2(-1) (F) x= 0 ; (0)2 = 2(0) (T) x= 1 ; (1)2 = 2(1) (F) F F x∈U F x2 = 2x ∃x[ x 2 = 2 x] ; U = { -1, 0, 1 } F
19. 19. F 31201 ก 4 WiLa 19 4. ∃x[ x + 1 = 1] U = R F F x∈U F x=o F x+1 = 1 ∃x[ x + 1 = 1] U= R F 5. ∀ x [ x+1 = x ] U = R F F x∈U F x=o F x+1 = x ∀ x [x+1 = x ] U = R F ก 91. F FF 1. ∀ x[ x+x = 2x ] ; u = {-2,-1,0,1,2} 2. ∃ x[ (x-1) (x+1) = x2 - 1 ] ; U = { -2,1,3,7} 3. ∃ x[2x2+3x+1 = 0] ; U = {-2,1,3,7} 4. ∀ x[x2+2x+1 = 0] ; u = {-2,1,3,7} 5. ∀ x[ x+3 < 5 ] ; U = R-2. F U = {-2,-1, 0, 1, 2} P(x) x ≥0 ; Q(x) x/4 R(x) x2-4 =0 F FF 1. ∀x[ P( x)∨ ~ R( x)] 2. → P(x)] ∃ x[ R(x) 3. ∀ x[ Q(x) ↔ P(x)]3. F U=R P(x) x ก ; Q(x) x ก F FF 1. ∀ x[P(x)] ∨ ∀ x [Q(x0] 2. ∀ x[P(x) → Q(x) ] 3. ∀ x(P(x) ∨ Q(x)] 4. ∀ x(P(x)]∧ ∀ x[Q(x)] 5. ∀ x[P(x) ∧Q(x)] ______________ ^__^ ______________
20. 20. F 31201 ก 4 WiLa 20F F F ก ก FU 1. F ∀x∀y [ p( x, y )] F ก F F x p(x,y) F กa U F F F ∀y  p ( a, y )    2. F ∀x∀y [ p( x, y )] F ก F กb U F F x p(x,y) F F F ∀y [ p(b, y)] 3. F ∀y∀x [ p( x, y )] F ก F F y p(x,y) F กa U F F F ∀x  p ( x, a )    4. F ∀y∀x [ p( x, y )] F ก F กb U F F y p(x,y) F F F ∀x [ p( x, b)] 5. F ∃x∃y [ p( x, y )] F ก F กa U F F a x p(x,y) F F F ∃y [ p(a, y )] 6. F ∃x∃y [ p( x, y )] F ก F F x p(x,y) F กa U F F F ∃y  p ( a, y )   7. F ∃y∃x [ p( x, y )] F ก F กa U F F a y p(x,y) F F F ∃x [ p( x, a)] 8. F ∃y∃x [ p( x, y )] F ก F F y p(x,y) F กa U F F F ∃x  p ( x, a )    9. F ∀x∃y [ p( x, y )] F ก F F x p(x,y) F กa U F F F ∃y  p ( a, y )   10. F ∀x∃y [ p( x, y )] F ก F กb U F F x p(x,y) F F F ∃y [ p(b, y )] 11. F ∀y∃x [ p( x, y )] F ก F F y p(x,y) F กa U F F F ∃x  p ( x, a )    12. F ∀y∃x [ p( x, y )] F ก F กb U F F y p(x,y) F F F ∃x [ p( x, b)] 13. F ∃x∀y [ p( x, y )] F ก F กa U F F x p(x,y) F F F ∀y [ p(a, y)] 14. F ∃x∀y [ p( x, y )] F ก F F x p(x,y) F กb U F F F ∀y  p ( b, y )    15. F ∃y∀x [ p( x, y )] F ก F กa U F F y p(x,y) F F F ∀x [ p( x, a)]
21. 21. F 31201 ก 4 WiLa 21 16. F ∃y∀x [ p( x, y )] F ก F F y p(x,y) F กb U F F F ∀x  p ( x, b )   1.10 ก ก F F ก ก ก ก F ก F ก ก F F ก FF F ก F F F ก ก F p(x),q(x) p(x,y) 1. ∀x[ p(x) ∧ q (x)] ≡ ∀x[ p(x)] ∧ ∀x[ q (x)] 2. ∃x[ p( x) ∨ q ( x)] ≡ ∃x[p( x)] ∨ ∃x[q ( x)] 3. ~ ∀x[ p(x)] ≡ ∃x[ ~ p(x)] 4. ~ ∃x[p( x)] ≡ ∀x[ ~ p( x)] 5. ~ ∀x∀y [ p( x, y)] ≡ ∃x∃y [ ~ p( x, y)] 6. ~ ∀x∃y [ p( x, y)] ≡ ∃x∀y [ ~ p( x, y)] 7. ~ ∃x∀y [ p( x, y)] ≡ ∀x∃y [ ~ p( x, y)]