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Logic Examples

Neja Cesnik Argumentation and Computing Inglés 66 fichas

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  1. 1If A U B = B, then B ⊂ AFalse
  2. 2If A U B = B, then A ⊂ BTrue
  3. 3If A ⊂ B, then A U B = BTrue
  4. 4If A ⊂ B, then A U B = AFalse
  5. 5A ∩ A = ATrue
  6. 6A ∩ B = B U AFalse
  7. 7A ∩ (B ∩ C) = (A ∩ B) ∩ CTrue
  8. 8A ∩ 0 = 0True
  9. 9If A ∩ B = B, then B ⊂ ATrue
  10. 10If A ∩ B = B, then A ⊂ BFalse
  11. 11If A ⊂ B, then A ∩ B = BFalse
  12. 12If A ⊂ B, then A ∩ B = ATrue
  13. 13If A ⊂ B, then A' ⊂ B'False
  14. 14If A ⊂ B, then B' ⊂ A'True
  15. 15Write truth tables for De Morgan's laws and express them as Venn diagramsCheck Tutorial 2
  16. 16- (P ʌ Q) ↔ (-P ʌ -Q)T2
  17. 17P ↔ -PT2
  18. 18P → -Q; -Q → R; P ⊢ R1. P → ¬Q A
    2. ¬Q → R A
    3. P A
    4. ¬Q 1,3 MPP
    5. R 2,4 MPP
  19. 19P → (Q → R); P; -R ⊢ Q1. P → (Q → R) A
    2. P A
    3. ¬R A
    4. Q → R 1, 2 MPP
    5. ¬Q 3,4 MTT
  20. 20P → -P ⊢ -P1. P → ¬P A
    2. P A
    3. ¬P 1, 2 MPP
    4. P ∧ ¬P 2, 3 ∧I
    5. ¬ P 2,4 RAA
  21. 21P → Q; P → -Q ⊢ -P1. P → Q A
    2. P → ¬Q A
    3. P A
    4. Q 1,3 MPP
    5. ¬Q 2,3 MPP
    6. Q ∧ ¬Q 4,5 ∧I
    7. ¬P 3,6 RAA
  22. 22P ʌ Q ⊢ Q ʌ P1. P ∧ Q A
    2. P &E
    3. Q &E
    4.Q∧P 3,2&I
  23. 23Q → R ⊢ (P ʌ Q) → (P ʌ R)1. Q → R A
    2. P ∧ Q A
    3.P 2 &E
    4.Q 2 &E
    5.R 1, 4 MPP
    6. P ∧ R 3,5 &I
    7. (P ∧ Q) → (P ∧ R) 2, 6 CP
  24. 24P ⊢ Q → (P ʌ Q)1. P A
    2. Q A
    3. P ∧ Q 1,2 &I
    4.Q→(P∧Q) 2,3CP
  25. 25Give two examples of a binary predicate and for each, state whether it is reflexive, symmetric or transitiveP3
  26. 26Rxy x recognises y, Universe of discourse all the people in the room. Is it reflexive, symmetric or transitive?R is reflexive, not symmetric, not transitive.
  27. 27Everyone in the room recognises someone in the room∀x, ∃y st Rxy
  28. 28There is someone in the room whom everyone recognises∃x st ∀y, Ryx
  29. 29There is someone x who recognises everyone who recognises x∃xst∀y(Ryx→Rxy)
  30. 30For every number there is greater number∀x,∃y(y>x)
  31. 31Every number other than 0 is greater than some number∀x(¬(x = 0) → ∃y(x>y))
  32. 320 is the one and only number having the property that no number is less than it∀x(¬∃y(y
  33. 33X is equal to y (without using =)¬(xy)
  34. 34X is unequal to y(xy)
  35. 35Some girl won all the prizes∃x(girl(x) ∧ ∀y(prize(y) → won(x, y)))
  36. 36Each of the prizes was won by a girl∀x(prize(x) → (∃y(girl(y) ∧ won(y, x))))
  37. 37If any of the brakes hold, the train will halt∀x((brake(x) ∧ hold(x)) → train will halt)
  38. 38Brand x doesn't remove any kind of stain∀y(Stain(y) → ¬BrandXRemoves(y))
  39. 39Brand x doesn't remove all kinds of stain∃y s.t. (Stain(y) ∧ ¬BrandXRemoves(y)) or ¬∀y(Stain(y) → ¬BrandXRemoves(y))
  40. 40Different types of reasoning i)Induction, not valid
  41. 41Different types of reasoning ii)Deduction, valid
  42. 42Different types of reasoning iii)Abduction, not valid
  43. 43Different types of reasoning 1./
  44. 44Different types of reasoning 2./
  45. 45Different types of reasoning 3./
  46. 46Different types of reasoning 4./
  47. 47Different types of reasoning 5./
  48. 48Different types of reasoning 6./
  49. 49P → Q; P├ Q1. P → Q A
    2. P A
    3. Q 1,2 MPP
  50. 50-Q → (-P → Q); -Q├ -P → Q1. -Q → (-P → Q) A
    2. -Q A
    3. -P → Q 1, 2 MPP
  51. 51P → Q; Q → R; P ├ R1. P → Q A
    2. Q → R A
    3. P A
    4. Q 1,3 MPP
    5. R 2,4 MPP
  52. 52P → Q; -Q├ -P1. P → Q A
    2. -Q A
    3. -P 1,2 MTT
  53. 53P → (Q → R); P, -R├ -Q1. P → (Q → R) A
    2. P A
    3. -R A
    4. Q → R 1,2 MPP
    5. -Q 3,4 MTT
  54. 54P → -Q; Q├ -P1. P → -Q A
    2. Q A
    3. - -Q 2 DN
    4. -P 1,3 MTT
  55. 55- P → Q; -Q├ P1. - P → Q A
    2. - Q A
    3. - - P 1,2 MTT
    4. P 3 DN
  56. 56P; Q├ P ∧ Q1. P A
    2. Q A
    3. P ∧ Q 1,2 &I
  57. 57P ∧ Q├ P1. P ∧ Q A
    2. P 1 &E
  58. 58P → Q; P → -Q├ -P1. P → Q A
    2. P → -Q A
    3. P A
    4. Q 1,3 MPP
    5. -Q 2,3 MPP
    6. Q ∧ -Q 4,5 &I
    7. -P 3,6 RAA
  59. 59Holmes can catch anyone who can catch Moriarty∀x (xCm → hCx)
  60. 60Holmes can catch anyone whom Moriarty can catch∀x (mCx → hCx)
  61. 61Holmes can catch anyone who can be caught by Moriarty∀x (mCx → hCx)
  62. 62If anyone can catch Moriarty, then Holmes can∃x (xCm) → hCm
  63. 63If everyone can catch Moriarty, then Holmes can∀x (xCm) → hCm
  64. 64Anyone who can catch Holmes can catch Moriarty∀x (xCh → xCm)
  65. 65Prove that: ∀x(Fx -> Gx), ∃x Fx├ ∃x Gx1. ∀x(Fx → Gx) (A)
    2. ∃x Fx (A)
    3. Fa (2, EE)
    4. Fa → Ga (1, UE)
    5. Ga (3,4, MPP)
    6. ∃x Gx (5,EI)
  66. 66∀x(Gx → Hx), ∃x (Fx ∧ Gx) ├ ∃x (Fx ∧ Hx)1.∀x(Gx → Hx) (A)
    2. ∃x (Fx ∧ Gx) (A)
    3. Fa ∧ Ga (2, EE)
    4. Ga → Ha (1, UE)
    5. Ga (3, ∧E)
    6. Ha (4,5, MPP)
    7. Fa (3, ∧E)
    8. Fa ∧ Ha (6,7 ∧I)
    9. ∃x (Fx ∧ Hx) (8, EI)

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Descripción
(Logic) Argumentation and Computing Fichas sobre Logic Examples, creado por Neja Cesnik el 30/04/2018.
Creado por
Neja Cesnik
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