Suggested Problems for Proof Designer
- Hypotheses: AB,
AC
Conclusion: ABC
- Hypotheses: AB
Conclusion: C \ BC
\ A
- Hypotheses: A \ BC
Conclusion: A \ CB
- Hypotheses: none
Conclusion: A \ (B \ C)(A \ B)C
- Hypotheses: none
Conclusion: A \ (BC)
= (A \ B)(A \
C)
- Hypotheses: none
Conclusion: A(BC)(AB)C
- Hypotheses: none
Conclusion: (AB) \
CA(B \ C)
- Hypotheses: A(B \
C) =
Conclusion: ABC
- Hypotheses: AB,
AC
Conclusion: BC
- Hypotheses: AB,
AC =
Conclusion: AB \ C
- Hypotheses: AB \
C, A ≠
Conclusion: BC
- Hypotheses: A \ B
C, AC
Conclusion: AB ≠
- Hypotheses: AB \
C
Conclusion: AC =
- Hypotheses: none
Conclusion: A \ C(A \ B)(B \ C)
- Hypotheses: ACBC, ACBC
Conclusion: AB
- Hypotheses: none
Conclusion: !AB(AB = B)
- Hypotheses: none
Conclusion: AB(A)(B)
- Hypotheses: none
Conclusion: (AB) = (A)(B)
- Hypotheses: none
Conclusion: (A)(B)(AB)
- Hypotheses: (A)(B)
= (AB)
Conclusion: ABBA
- Hypotheses:
x(xAxA)
Conclusion: x(x(A)x(A))
- Hypotheses: AF
Conclusion: AF
- Hypotheses: AF
Conclusion: UFA
- Hypotheses: FG
Conclusion: FG
- Hypotheses: FG
Conclusion: UGUF
- Hypotheses: none
Conclusion: (FG) = (F)(G)
- Hypotheses: none
Conclusion: (FG)(F)(G)
- Hypotheses: none
Conclusion: U(FG) =
(UF)(UG)
- Hypotheses: none
Conclusion: A(F) = {AX | XF}
- Hypotheses: AU
Conclusion: A(UF) =
U{AX |
XF}
- Hypotheses: none
Conclusion: U \ F =
U{U \ X | XF}
- Hypotheses: AU
Conclusion: A \ (UF) = {A \ X | XF}
- Hypotheses: none
Conclusion: F \ G(F \ G)
- Hypotheses: none
Conclusion: (F \ G)F
\ GFG(FG)
- Hypotheses:
AFBG(ABH)
Conclusion: (F)GH
- Hypotheses: none
Conclusion: F(F)
- Hypotheses: none
Conclusion: A = (A)
- Hypotheses: none
Conclusion: UF(U){(X) | XF}
- Hypotheses: none
Conclusion: {X \ A |
XF}{XF | XA}
- Hypotheses: none
Conclusion: (F)(G) = AFBG(AB = )
- Hypotheses: none
Conclusion: {(X) | XF}(F)
- Hypotheses: none
Conclusion: (U){(X) | XF} = (UF)
- Hypotheses: {(X) | XF} = (F)
Conclusion: AFBF(BA)
- Hypotheses: F(F = AAF)
Conclusion: x(A =
{x})
- Hypotheses: none
Conclusion: (A \ B) \
((A) \ (B)) = {}
- Hypotheses: none
Conclusion: A(BC) = (AB)(AC)
- Hypotheses: none
Conclusion: A(BC) = (AB)(AC)
- Hypotheses: none
Conclusion: (AB)C = (AC)(BC)
- Hypotheses: ABB
Conclusion: AB
- Hypotheses: none
Conclusion: AB(AC)(BC)
- Hypotheses: none
Conclusion: A(AB) = A \ B
- Hypotheses: none
Conclusion: A(AB) = B \ A
- Hypotheses: none
Conclusion: (AB)C = A(BC)
- Hypotheses: none
Conclusion: AA =
- Hypotheses: AC =
BC
Conclusion: A = B
- Hypotheses: none
Conclusion: !AB(AB = B)
- Hypotheses: none
Conclusion: AB!C(AC
= B)
- Hypotheses: none
Conclusion: UA(AU)
- Hypotheses: none
Conclusion: (RS)-1 = S-1R-1
- Hypotheses: none
Conclusion: (RS)T = R(ST)
- Hypotheses: ST
Conclusion: RSRT
- Hypotheses: none
Conclusion: (ST)R(SR)(TR)
- Hypotheses: none
Conclusion: (ST)R = (SR)(TR)
- Hypotheses: none
Conclusion: (SR) \
(TR)(S \ T)R