Suggested Problems for Proof Designer

  1. Hypotheses: AB, AC
    Conclusion: ABC
  2. Hypotheses: AB
    Conclusion: CBCA
  3. Hypotheses: ABC
    Conclusion: ACB
  4. Hypotheses: none
    Conclusion: A∖(BC)⊆(AB)∪C
  5. Hypotheses: none
    Conclusion: A∖(BC)=(AB)∪(AC)
  6. Hypotheses: none
    Conclusion: A∩(BC)⊆(AB)∪C
  7. Hypotheses: none
    Conclusion: (AB)∖CA∪(BC)
  8. Hypotheses: A∩(BC)=∅
    Conclusion: ABC
  9. Hypotheses: AB, AC
    Conclusion: BC
  10. Hypotheses: AB, AC=∅
    Conclusion: ABC
  11. Hypotheses: ABC, A≠∅
    Conclusion: BC
  12. Hypotheses: ABC, AC
    Conclusion: AB≠∅
  13. Hypotheses: ABC
    Conclusion: AC=∅
  14. Hypotheses: none
    Conclusion: AC⊆(AB)∪(BC)
  15. Hypotheses: ACBC, ACBC
    Conclusion: AB
  16. Hypotheses: none
    Conclusion: ∃!AB(AB=B)
  17. Hypotheses: none
    Conclusion: AB↔𝒫(A)⊆𝒫(B)
  18. Hypotheses: none
    Conclusion: 𝒫(AB)=𝒫(A)∩(B)
  19. Hypotheses: none
    Conclusion: 𝒫(A)∪𝒫(B)⊆𝒫(AB)
  20. Hypotheses: 𝒫(A)∪𝒫(B)=𝒫(AB)
    Conclusion: ABBA
  21. Hypotheses: x(xAxA)
    Conclusion: x(x∈𝒫(A)→x⊆𝒫(A))
  22. Hypotheses: AF
    Conclusion: A⊆∪F
  23. Hypotheses: AF
    Conclusion: U∩∩FA
  24. Hypotheses: FG
    Conclusion: F⊆∪G
  25. Hypotheses: FG
    Conclusion: U∩∩GU∩∩F
  26. Hypotheses: none
    Conclusion: ∪(FG)=(∪F)∪(∪G)
  27. Hypotheses: none
    Conclusion: ∪(FG)⊆(∪F)∩(∪G)
  28. Hypotheses: none
    Conclusion: U∩∩(FG)=(U∩∩F)∩(U∩∩G)
  29. Hypotheses: none
    Conclusion: A∩(∪F)=∪{AX|XF}
  30. Hypotheses: AU
    Conclusion: A∪(U∩∩F)=U∩∩{AX|XF}
  31. Hypotheses: none
    Conclusion: U∖∪F=U∩∩{UX|XF}
  32. Hypotheses: AU
    Conclusion: A∖(U∩∩F)=∪{AX|XF}
  33. Hypotheses: none
    Conclusion: F∖∪G⊆∪(FG)
  34. Hypotheses: none
    Conclusion: ∪(FG)⊆∪F∖∪G→∪F∩∪G⊆∪(FG)
  35. Hypotheses: AFBG(ABH)
    Conclusion: (∪F)∩∩G⊆∪H
  36. Hypotheses: none
    Conclusion: F⊆𝒫(∪F)
  37. Hypotheses: none
    Conclusion: A=∪𝒫(A)
  38. Hypotheses: none
    Conclusion: U∩∩F∈𝒫(U)∩∩{𝒫(X)|XF}
  39. Hypotheses: none
    Conclusion: ∪{XA|XF}⊆∪{XF|XA}
  40. Hypotheses: none
    Conclusion: (∪F)∩(∪G)=∅↔∀AFBG(AB=∅)
  41. Hypotheses: none
    Conclusion: ∪{𝒫(X)|XF}⊆𝒫(∪F)
  42. Hypotheses: none
    Conclusion: 𝒫(U)∩∩{𝒫(X)|XF}=𝒫(U∩∩F)
  43. Hypotheses: ∪{𝒫(X)|XF}=𝒫(∪F)
    Conclusion: AFBF(BA)
  44. Hypotheses: F(∪F=AAF)
    Conclusion: x(A={x})
  45. Hypotheses: none
    Conclusion: 𝒫(AB)∖(𝒫(A)∖𝒫(B))={∅}
  46. Hypotheses: none
    Conclusion: A×(BC)=(A×B)∩(A×C)
  47. Hypotheses: none
    Conclusion: A×(BC)=(A×B)∪(A×C)
  48. Hypotheses: none
    Conclusion: (AB)∩C=(AC)△(BC)
  49. Hypotheses: ABB
    Conclusion: AB
  50. Hypotheses: none
    Conclusion: AB⊆(AC)∪(BC)
  51. Hypotheses: none
    Conclusion: A△(AB)=AB
  52. Hypotheses: none
    Conclusion: A△(AB)=BA
  53. Hypotheses: none
    Conclusion: (AB)△C=A△(BC)
  54. Hypotheses: none
    Conclusion: AA=∅
  55. Hypotheses: AC=BC
    Conclusion: A=B
  56. Hypotheses: none
    Conclusion: ∃!AB(AB=B)
  57. Hypotheses: none
    Conclusion: AB∃!C(AC=B)
  58. Hypotheses: none
    Conclusion: ¬∃UA(AU)
  59. Hypotheses: none
    Conclusion: (RS)−1=S−1R−1
  60. Hypotheses: none
    Conclusion: (RS)○T=R○(ST)
  61. Hypotheses: ST
    Conclusion: RSRT
  62. Hypotheses: none
    Conclusion: (ST)○R⊆(SR)∩(TR)
  63. Hypotheses: none
    Conclusion: (ST)○R=(SR)∪(TR)
  64. Hypotheses: none
    Conclusion: (SR)∖(TR)⊆(ST)○R