Case Index¶
Below you'll find all of the cases in pyetr.cases, and their associated views. You can use this page as an index of the current cases.
e1¶
description:
Example 1, p61:
P1 Either Jane is kneeling by the fire and she is looking at the TV or else
Mark is standing at the window and he is peering into the garden.
P2 Jane is kneeling by the fire.
C Jane is looking at the TV.
v[0]: {KneelingByTheFire(Jane())LookingAtTV(Jane()),PeeringIntoTheGarden(Mark())StandingAtTheWindow(Mark())}
v[1]: {KneelingByTheFire(Jane())}
c (Conclusion): {LookingAtTV(Jane())}
test(verbose=False): Method used to test the example
e2¶
description:
Example 2, p62:
P1 There is at least an ace and a queen, or else at least a king and a ten.
P2 There is a king.
C There is a ten.
v[0]: {A()Q(),K()T()}
v[1]: {K()}
c (Conclusion): {T()}
test(verbose=False): Method used to test the example
e3¶
description:
Example 3, p63:
P1 There is at least an ace and a king or else there is at least a queen and
a jack.
P2 There isn't an ace.
C There is a queen and a jack.
v[0]: {Ace()King(),Jack()Queen()}
v[1]: {~Ace()}
c (Conclusion): {Jack()Queen()}
test(verbose=False): Method used to test the example
e5ii¶
description:
Example 5, p72, part ii
v[0]: {p1()q1(),r1()s1()}
v[1]: {p2()q2(),r2()s2()}
c (Conclusion): {p1()p2()q1()q2(),p1()q1()r2()s2(),p2()q2()r1()s1(),r1()r2()s1()s2()}
test(verbose=False): Method used to test the example
e5iii¶
description:
Example 5, p72, part iii
v[0]: {p1()q1(),r1()s1()}
v[1]: {}
c (Conclusion): {}
test(verbose=False): Method used to test the example
e5iv¶
description:
Example 5, p72, part iv
v[0]: {p1()q1(),r1()s1()}
v[1]: {0}
c (Conclusion): {p1()q1(),r1()s1()}
test(verbose=False): Method used to test the example
e5v¶
description:
Example 5, p72, part v
v[0]: {0}
v[1]: {p1()q1(),r1()s1()}
c (Conclusion): {p1()q1(),r1()s1()}
test(verbose=False): Method used to test the example
e6¶
description:
Example 6, p72
There is an Ace and a King = (There is an Ace) x (There is a king)
v[0]: {a()}
v[1]: {k()}
c (Conclusion): {a()k()}
test(verbose=False): Method used to test the example
e7¶
description:
Example 7, p73
There is an Ace or there is a king = (There is an Ace) + (There is a king)
v[0]: {a()}
v[1]: {k()}
c (Conclusion): {a(),k()}
test(verbose=False): Method used to test the example
e8¶
description:
Example 8, p74
P1 There is an ace and a queen, or else there is a king and a ten
P2 There is a king
C There is a ten (and a king)
v[0]: {a()q(),k()t()}
v[1]: {k()}
c (Conclusion): {t()}
test(verbose=False): Method used to test the example
e10¶
description:
Example 10, p76
P1 There is a king.
P2 There is at least an ace and a queen, or else at least a king and a ten.
C There is a king (reversed premises blocking illusory inference).
v[0]: {K()}
v[1]: {A()Q(),K()T()}
c (Conclusion): {K()}
test(verbose=False): Method used to test the example
e11¶
description:
Example 11, p77
P1 Either John smokes or Mary smokes.
P2 Supposing John smokes, John drinks.
P3 Supposing Mary smokes, Mary eats.
C Either John smokes and drinks or Mary smokes and drinks.
v[0]: {Smokes(j()),Smokes(m())}
v[1]: {Drinks(j())}^{Smokes(j())}
v[2]: {Eats(m())}^{Smokes(m())}
c (Conclusion): {Drinks(j())Smokes(j()),Eats(m())Smokes(m())}
test(verbose=False): Method used to test the example
e12i¶
description:
Example 12i, p78
ItisnotthecasethatPorQorR
v[0]: {P(),Q(),R()}
c (Conclusion): {~P()~Q()~R()}
test(verbose=False): Method used to test the example
e12ii¶
description:
Example 12ii, p78
ItisnotthecasethatPandQandR
v[0]: {P()Q()R()}
c (Conclusion): {~P(),~Q(),~R()}
test(verbose=False): Method used to test the example
e12iii¶
description:
Example 12iii, p79
It is not the case that, supposing S, ((P and Q) or R)
v[0]: {P()Q(),R()}^{S()}
c (Conclusion): {S()~P()~R(),S()~Q()~R()}
test(verbose=False): Method used to test the example
e13¶
description:
Example 13, p80
P1 There is an ace and a king or a queen and a jack.
P2 There isn't an ace.
C There is a queen and a jack.
v[0]: {IsAce()IsKing(),IsJack()IsQueen()}
v[1]: {~IsAce()}
c (Conclusion): {IsJack()IsQueen()}
test(verbose=False): Method used to test the example
e14_1¶
description:
Example 14-1, p81
Factor examples
v[0]: {P()Q(),P()R()}
v[1]: {P()}
c (Conclusion): {Q(),R()}
test(verbose=False): Method used to test the example
e14_2¶
description:
Example 14-2, p81
Factor examples
v[0]: {P()Q()S(),P()R(),P()R()S()}
v[1]: {P()}^{S()}
c (Conclusion): {P()R(),Q()S(),R()S()}
test(verbose=False): Method used to test the example
e14_3¶
description:
Example 14-3, p81
Factor examples
v[0]: {P()R(),P()S(),Q()R(),Q()S()}
v[1]: {P(),Q()}
c (Conclusion): {R(),S()}
test(verbose=False): Method used to test the example
e14_6¶
description:
Example 14-6, p81
Factor examples
v[0]: {P()R(),Q()S()}
v[1]: {P(),Q(),T()}
c (Conclusion): {P()R(),Q()S()}
test(verbose=False): Method used to test the example
e14_7¶
description:
Example 14-7, p81
Factor examples
v[0]: {P(),P()R(),Q()S()}
v[1]: {P(),Q()}
c (Conclusion): {0,R(),S()}
test(verbose=False): Method used to test the example
e15¶
description:
Example 15, p82
P1 There is an ace and a jack and a queen, or else there is an eight and a ten and a four, or else there is an ace.
P2 There is an ace and a jack, and there is an eight and a ten.
P3 There is not a queen.
C There is a four
v[0]: {Ace(),Ace()Jack()Queen(),Eight()Four()Ten()}
v[1]: {Ace()Eight()Jack()Ten()}
v[2]: {~Queen()}
c (Conclusion): {Four()}
test(verbose=False): Method used to test the example
e16¶
description:
Example 16, p83
P1 There is a ten and an eight and a four, or else there is a jack and a king and a queen, or else there is an ace.
P2 There isn't a four.
P3 There isn't an ace.
v[0]: {Ace(),Eight()Four()Ten(),Jack()King()Queen()}
v[1]: {~Four()}
v[2]: {~Ace()}
c (Conclusion): {Jack()King()Queen()}
test(verbose=False): Method used to test the example
e17¶
description:
Example 17, p83
P1 There is a king in the hand and there is not an ace in the hand, or else there is an ace in the hand and there is not a king in the hand.
P2 There is a king in the hand.
C There isn't an ace in the hand.
v[0]: {Ace()~King(),King()~Ace()}
v[1]: {King()}
c (Conclusion): {~Ace()}
test(verbose=False): Method used to test the example
e19¶
description:
Example 19, p84
Suppose test
v[0]: {0}
v[1]: {~N()}
c (Conclusion): {~N()}^{~N()}
test(verbose=False): Method used to test the example
e20¶
description:
Example 20, p85
P1 Either there is a king in the hand or a queen in the hand.
P2 On the supposition that there is a king, Mary wins.
P3 On the supposition that there is a queen, Bill wins.
C Either Mary wins or Bill wins.
v[0]: {King(),Queen()}
v[1]: {Win(mary())}^{King()}
v[2]: {Win(bill())}^{Queen()}
c (Conclusion): {Win(bill()),Win(mary())}
test(verbose=False): Method used to test the example
e21¶
description:
Example 21, p86
Any view Δ^{0} = [Δ^{0}]ᶰ can be derived from the absurd view
v[0]: {r1()s1()}
c (Conclusion): {~r1(),~s1()}
test(verbose=False): Method used to test the example
e22¶
description:
Example 22, p87
It is not the case that A and B and C
v[0]: {a()b()c()}
v[1]: {a()}
v[2]: {b()}
v[3]: {c()}
v[0]: {~a(),~b(),~c()}
v[1]: {a()b()~c(),a()c()~b(),a()~b()~c(),b()c()~a(),b()~a()~c(),c()~a()~b(),~a()~b()~c()}
test(verbose=False): Method used to test the example
e23_with_inquire¶
description:
Example 23, p88, with inquire step
P1 Either Jane is kneeling by the fire and she is looking at the TV or else Mark is
standing at the window and he is peering into the garden.
P2 Jane is kneeling by the fire
C Jane is looking at the TV
v[0]: {K()L(),P()S()}
v[1]: {K()}
v[0]: {K()L(),K()P()S(),P()S()~K()}
v[1]: {K()L(),K()P()S()}
test(verbose=False): Method used to test the example
e23_without_inquire¶
description:
Example 23, p88, without inquire step
P1 Either Jane is kneeling by the fire and she is looking at the TV or else Mark is
standing at the window and he is peering into the garden.
P2 Jane is kneeling by the fire
C Jane is looking at the TV
v[0]: {K()L(),P()S()}
v[1]: {K()}
v[0]: {K()L(),P()S()}
v[1]: {K()L()}
test(verbose=False): Method used to test the example
e24¶
description:
Example 24, p89
P1 There is an ace
C There is an ace or a queen
v[0]: {a()}
v[1]: {q()}
v[2]: {~q()}
v[3]: {a(),q()}
v[0]: {a()q(),a()~q()}
v[1]: {a(),q()}
test(verbose=False): Method used to test the example
e25i¶
description:
Example 25i, p89
v[0]: {p()q(),p()r()}
v[1]: {p()}
c (Conclusion): {p()}
test(verbose=False): Method used to test the example
e25ii¶
description:
Example 25ii, p89
v[0]: {p()q(),p()r()}
v[1]: {q()}
c (Conclusion): {0,q()}
test(verbose=False): Method used to test the example
e25iii¶
description:
Example 25iii, p89
v[0]: {p()q(),p()r(),s(),t()}
v[1]: {p(),s()}
c (Conclusion): {0,p(),s()}
test(verbose=False): Method used to test the example
e25iv¶
description:
Example 25iv, p89
v[0]: {p()q(),p()r(),s(),t()}
v[1]: {p(),s(),t()}
c (Conclusion): {p(),s(),t()}
test(verbose=False): Method used to test the example
e25v¶
description:
Example 25v, p89
v[0]: {p()q()s(),p()r()s()}
v[1]: {p()}^{s()}
c (Conclusion): {p()}
test(verbose=False): Method used to test the example
e25vi¶
description:
Example 25vi, p89
v[0]: {p()q()s(),p()r()s()}
v[1]: {p()}^{t()}
c (Conclusion): {0}
test(verbose=False): Method used to test the example
e26¶
description:
Example 26, p90
P1 Either John plays and wins, or Mary plays, or Bill plays
C Supposing John plays, John wins
v[0]: {Play(B()),Play(J())Win(J()),Play(M())}
v[1]: {Play(J())}
v[2]: {Win(J())}^{Play(J())}
v[0]: {Play(J())Win(J())}^{Play(J())}
v[1]: {Win(J())}^{Play(J())}
test(verbose=False): Method used to test the example
e26_does_it_follow¶
description:
Example 26, p90
P1 Either John plays and wins, or Mary plays, or Bill plays
C Supposing John plays, John wins
v[0]: {Play(B()),Play(J())Win(J()),Play(M())}
v[1]: {Play(J())}
v[2]: {Win(J())}^{Play(J())}
v[0]: {Play(J())Win(J())}^{Play(J())}
v[1]: {Win(J())}^{Play(J())}
test(verbose=False): Method used to test the example
e28¶
description:
Example 28, p96
P1 Is there a tiger?
P2 Supposing there is a tiger, there is orange fur.
P3 There is orange fur.
C There is a tiger.
v[0]: {Tiger(),~Tiger()}
v[1]: {Orange()Tiger()}^{Tiger()}
v[2]: {Orange()}
c (Conclusion): {Orange()Tiger()}
test(verbose=False): Method used to test the example
e32_1¶
description:
Example 32-1, p107
P1 If P then Q.
P2 P
C Q
v[0]: {P()Q()}^{P()}
v[1]: {P()}
c (Conclusion): {Q()}
test(verbose=False): Method used to test the example
e32_2¶
description:
Example 32-2, p107
P1 P
P2 If P then Q.
C Q
v[0]: {P()}
v[1]: {P()Q()}^{P()}
c (Conclusion): {Q()}
test(verbose=False): Method used to test the example
e33¶
description:
Example 33, p108
P1 If the card is red then the number is even.
P2 The number is even.
C The card is red
v[0]: {E()R()}^{R()}
v[1]: {E()}
c (Conclusion): {R()}
test(verbose=False): Method used to test the example
e40i¶
description:
Example 40, p119
(P0 Shapes at the bottom of the card are mutually exclusive)
P1 If there is a circle at the top of the card, then there is a
square on the bottom.
P2 There is a triangle on the bottom
C Falsum
v[0]: {CircleB()~SquareB()~TriangleB(),SquareB()~CircleB()~TriangleB(),TriangleB()~CircleB()~SquareB()}
v[1]: {CircleT()SquareB()}^{CircleT()}
v[2]: {TriangleB()}
c (Conclusion): {}
test(verbose=False): Method used to test the example
e40ii¶
description:
Example 40, p119-p120
(P0 Shapes at the bottom of the card are mutually exclusive)
P1 If there is a circle at the top of the card, then there is a
square on the bottom.
P2 There is a triangle on the bottom
C Falsum
The reader diverges from the default procedure,
and deposes the conditional premise, and switches the premise
order.
v[0]: {CircleB()~SquareB()~TriangleB(),SquareB()~CircleB()~TriangleB(),TriangleB()~CircleB()~SquareB()}
v[1]: {TriangleB()}
v[2]: {CircleT()SquareB()}^{CircleT()}
c (Conclusion): {TriangleB()~CircleB()~CircleT()~SquareB()}
test(verbose=False): Method used to test the example
e41¶
description:
Example 41, p121
P1 P only if Q.
P2 Not Q.
C Not P.
v[0]: {~P()~Q()}^{~Q()}
v[1]: {~Q()}
c (Conclusion): {~P()}
test(verbose=False): Method used to test the example
e42¶
description:
Example 42, p122
P1 There is a circle at the top of the card only if there is a square
at the bottom.
P2 There is not a square at the bottom
C There is not a circle at the top
v[0]: {~CircleT()~SquareB()}^{~SquareB()}
v[1]: {~SquareB()}
c (Conclusion): {~CircleT()}
test(verbose=False): Method used to test the example
e44_1¶
description:
Example 44-1, p123
P1 The chair is saleable if and only if it is inelegant.
P2 The chair is elegant if and only if it is stable.
P3 The chair is saleable or it is stable, or both.
C The chair is saleable elegant and stable.
v[0]: {Elegant(c())Saleable(c()),~Elegant(c())~Saleable(c())}
v[1]: {Elegant(c())Stable(c()),~Elegant(c())~Stable(c())}
v[2]: {Elegant(c())Saleable(c()),Saleable(c()),Stable(c())}
c (Conclusion): {Elegant(c())Saleable(c())Stable(c())}
test(verbose=False): Method used to test the example
e45¶
description:
Example 45, p125
It is possible that Steven is in Madrid and it is possible that Emma is in
Berlin.
Therefore it is possible that Steven is in Madrid and that Emma is in Berlin.
v[0]: {0,M()}
v[1]: {0,B()}
v[2]: {0,B()M()}
v[0]: {0,B(),B()M(),M()}
v[1]: {0,B()M()}
test(verbose=False): Method used to test the example
e46i¶
description:
Example 46, p126
P1 Pat is here then Viv is here
P2 Mo is here or else Pat is here, but not both
C No
v[0]: {P()V()}^{P()}
v[1]: {M()~P(),P()~M()}
v[2]: {0,M()V()}
v[0]: {M()~P(),P()V()~M()}
v[1]: {0}
test(verbose=False): Method used to test the example
e46ii¶
description:
Example 46, part ii, p126
If we had a view{VMR,VMS, T} and applied [{vm, 0}]Q we would get [{vm, 0}]
v[0]: {M()R()V(),M()S()V(),T()}
v[1]: {0,M()V()}
c (Conclusion): {0,M()V()}
test(verbose=False): Method used to test the example
e47¶
description:
Example 47, p129
P1: Some thermotogum stains gram-negative
P2: Maritima is a thermotogum
C: Maritima stains gram negative
v[0]: ∃x {StainsGramNegative(x)Thermotogum(x*)}
v[1]: {Thermotogum(Maritima()*)}
c (Conclusion): {StainsGramNegative(Maritima())}
test(verbose=False): Method used to test the example
e48¶
description:
Example 48, p130
P1 Some dictyoglomus is thermophobic.
P2 Turgidum is not a dictyoglomus.
C Truth
v[0]: ∃x {D(x*)T(x)}
v[1]: {~D(Turgidum()*)}
c (Conclusion): {0}
test(verbose=False): Method used to test the example
e49¶
description:
Example 49, p130
P1 Either there is an ace in Mary's hand and some other player has a king,
or else there is a queen in John's hand and some other player has a jack.
P2 Sally has a king
C Truth
v[0]: ∃x ∃y {Ace(Mary())King(x),Jack(y)Queen(John())}
v[1]: {King(Sally())}
c (Conclusion): {0}
test(verbose=False): Method used to test the example
e50_part1¶
description:
Example 50, part1, p131
Jack is looking at Sally, but Sally is looking at George. Jack is married, but George is
not. Is the married person looking at an unmarried person?
(A) Yes
(B) No
(C) Cannot be determined
v[0]: {L(j(),s())L(s(),g())}
v[1]: {M(j()*)~M(g()*)}
v[2]: {}
v[3]: ∃a ∃b {L(a,b)M(a*)~M(b*)}
v[0]: {L(j(),s())L(s(),g())M(j()*)~M(g()*)}
v[1]: {0}
test(verbose=False): Method used to test the example
e50_part2¶
description:
Example 50, part2, p131
Jack is looking at Sally, but Sally is looking at George. Jack is married, but George is
not. Is the married person looking at an unmarried person?
(A) Yes
(B) No
(C) Cannot be determined
v[0]: {L(j(),s())L(s(),g())}
v[1]: {M(j())~M(g())}
v[2]: {M(s())}
v[3]: ∃a ∃b {L(a,b)M(a*)~M(b*)}
c (Conclusion): ∃a ∃b {L(a,b)M(a*)~M(b*)}
g1 (Another View): {L(j(),s())L(s(),g())M(j())M(s())~M(g()),L(j(),s())L(s(),g())M(j())~M(g())~M(s())}
g2 (Another View): {L(j(),s())L(s(),g())M(j()*)M(s())~M(g()*),L(j(),s())L(s(),g())M(j()*)~M(g()*)~M(s()*)}
test(verbose=False): Method used to test the example
e50_part2_arbs¶
description:
Duplicate of e50, uses arb objects, some changes
v[0]: ∃g ∃j ∃s {L(j,s)L(s,g)M(j)~M(g)}
v[1]: ∃s {M(s)}
v[2]: ∃a ∃b {L(a,b)M(a*)~M(b*)}
c (Conclusion): ∃a ∃b {L(a,b)M(a*)~M(b*)}
g1 (Another View): ∃g ∃j ∃s {L(j,s)L(s,g)M(j)M(s)~M(g),L(j,s)L(s,g)M(j)~M(g)~M(s)}
g2 (Another View): ∃g ∃j ∃s {L(j,s)L(s,g)M(j*)M(s)~M(g*),L(j,s)L(s,g)M(j*)~M(g*)~M(s*)}
test(verbose=False): Method used to test the example
e51¶
description:
Example 51, p131
P1: Every archaeon has a nucleus
P2: Halobacterium is an archaeon
C: Halobacterium is an archaeon and has a nucleus
v[0]: ∀x {HasNucleus(x)IsArchaeon(x*)}^{IsArchaeon(x*)}
v[1]: {IsArchaeon(Halobacterium()*)}
c (Conclusion): {HasNucleus(Halobacterium())IsArchaeon(Halobacterium()*)}
test(verbose=False): Method used to test the example
e52¶
description:
Example 52, p132
P1 All Fs G.
P2 John Gs.
C John Fs and Gs.
v[0]: ∀x {F(x)G(x*)}^{F(x)}
v[1]: {G(John()*)}
c (Conclusion): {F(John())G(John()*)}
test(verbose=False): Method used to test the example
e53¶
description:
Example 53, p132 & p175
P All A are B.
C All B are A.
v[0]: ∀x {A(x)B(x)}^{A(x)}
v[1]: ∀x {B(x)}
v[2]: ∀x {A(x)B(x)}^{B(x)}
c (Conclusion): ∀x {A(x)B(x)}^{B(x)}
test(verbose=False): Method used to test the example
e53_does_it_follow¶
description:
Example 53, p132 & p175
P All A are B.
C All B are A.
v[0]: ∀x {A(x)B(x)}^{A(x)}
v[1]: ∀x {B(x)}
v[2]: ∀x {A(x)B(x)}^{B(x)}
c (Conclusion): ∀x {A(x)B(x)}^{B(x)}
test(verbose=False): Method used to test the example
e54¶
description:
Example 54, p133
P1 Sharks attack bathers.
P2 Whitey is a shark.
C Whitey attacks bathers.
v[0]: ∀x {0,Attack(x)Shark(x*)}^{Shark(x*)}
v[1]: {Shark(Whitey()*)}
c (Conclusion): {Attack(Whitey())Shark(Whitey()*)}
test(verbose=False): Method used to test the example
e56_default_inference¶
description:
Example 56, p134
P1: Every professor teaches some student
P2: Every student reads some book
C: Every professor teaches some student who reads some book
v[0]: ∀x ∃y {Professor(x)Student(y*)Teaches(x,y)}^{Professor(x)}
v[1]: ∀z ∃w {Book(w)Reads(z,w)Student(z*)}^{Student(z*)}
c (Conclusion): ∃b ∃y {0,Book(b)Reads(y,b)}
test(verbose=False): Method used to test the example
e56_basic_step¶
description:
Example 56, p134
P1: Every professor teaches some student
P2: Every student reads some book
C: Every professor teaches some student who reads some book
v[0]: ∀x ∃y {Professor(x)Student(y*)Teaches(x,y)}^{Professor(x)}
v[1]: ∀z ∃w {Book(w)Reads(z,w)Student(z*)}^{Student(z*)}
c (Conclusion): ∀a ∃b ∃c {Book(c)Professor(a)Reads(b,c)Student(b*)Teaches(a,b),~Professor(a)}
test(verbose=False): Method used to test the example
e57¶
description:
Example 57, p134
P1 All B are A.
P2 Some C are B.
C Some C are A.
v[0]: ∀x {A(x)B(x*)}^{B(x*)}
v[1]: ∃x {B(x*)C(x)}
c (Conclusion): ∃y {A(y)B(y*)C(y)}
test(verbose=False): Method used to test the example
e58_reversed¶
description:
Example 58 reversed, based on p135
P1 All C are B.
P2 Some B are A.
C Some C are A.
v[0]: ∀y {B(y*)C(y)}^{C(y)}
v[1]: ∃x {A(x)B(x*)}
c (Conclusion): ∃y {A(y)B(y*)C(y)}
test(verbose=False): Method used to test the example
e61¶
description:
Example 61, p166
P1 All dogs bite some man
P2 John is a man
C All dogs bite John
v[0]: ∀x ∃a {B(x,a)D(x)M(a*),~D(x)}
v[1]: {M(j()*)}
c (Conclusion): ∀x ∃a {B(x,a)D(x)M(a*)M(j()*),M(j()*)~D(x)}
test(verbose=False): Method used to test the example
e62¶
description:
Example 62, p176
v[0]: {D(b())~S(n()*),D(m())S(j()*)T(n()),L(n(),m())S(m()*)}
v[1]: ∃a {S(a*)}
c (Conclusion): {0,S(j()*),S(m()*)}
test(verbose=False): Method used to test the example
e63¶
description:
Example 63, p176
v[0]: {D(n()*)S(j()*),D(n()*)T(j())~D(j()*)}
v[1]: ∃a {D(a*)}
c (Conclusion): {D(n()*)}
test(verbose=False): Method used to test the example
e63_modified¶
description:
Example 63, p176
v[0]: ∀x ∃y {D(f(y,x)*)T(j())~D(j()*),D(n()*)S(j()*)}
v[1]: ∃a {D(a*)}
c (Conclusion): ∀x ∃y {D(f(y,x)*),D(n()*)}
test(verbose=False): Method used to test the example
e64i¶
description:
Example 64, p189, p223
A device has been invented for screening a population for a disease known as psylicrapitis.
The device is a very good one, but not perfect. If someone is a sufferer, there is a 90% chance
that he will recorded positively. If he is not a sufferer, there is still a 1% chance that he will
be recorded positively.
Roughly 1% of the population has the disease. Mr Smith has been tested, and the result is positive.
What is the chance that he is in fact a sufferer?
v[0]: ∀x {90.0=* S(x*)T(x*),S(x*)~T(x)}^{S(x*)}
v[1]: ∀x {1.0=* T(x)~S(x*),~S(x*)~T(x)}^{~S(x*)}
v[2]: {T(Smith()*)}
v[3]: {S(Smith())}
c (Conclusion): {0,90.0=* S(Smith()*)}
test(verbose=False): Method used to test the example
e64ii¶
description:
Example 64, p189, p223
A device has been invented for screening a population for a disease known as psylicrapitis.
The device is a very good one, but not perfect. If someone is a sufferer, there is a 90% chance
that he will recorded positively. If he is not a sufferer, there is still a 1% chance that he will
be recorded positively.
Roughly 1% of the population has the disease. Mr Smith has been tested, and the result is positive.
What is the chance that he is in fact a sufferer?
v[0]: ∀x {90.0=* P(x)S(x*)T(x*),P(x)S(x*)~T(x)}^{P(x)S(x*)}
v[1]: ∀x {1.0=* P(x)T(x)~S(x*),P(x)~S(x*)~T(x)}^{P(x)~S(x*)}
v[2]: ∀x {1.0=* P(x)S(x*),P(x)~S(x)}^{P(x)}
v[3]: {P(Smith())T(Smith()*)}
v[4]: {S(Smith())}
c (Conclusion): {90.0=* S(Smith()*)}
test(verbose=False): Method used to test the example
e65¶
description:
Example 65, p190, p224
(Base-rate neglect with doctors and realistic disease) Imagine you conduct
a screening using the Hemoccult test in a certain region. For symptom-free
people over 50 years old who participate in screening using the Hemoccult test,
the following information is available for this region.
The probability that one of these people has colorectal cancer is 0.3%. If a
person has colorectal cancer, the probability is 50 that he will have a positive
Hemoccult test. If a person does not have a colorectal cancer, the probability is
3% that he will still have a positive Hemoccult test in your screening. What is
the probability that this person actually has colorectal cancer?
v[0]: ∀x {0.3=* C(x)P(x*),P(x*)~C(x)}^{P(x*)}
v[1]: ∀x {50.0=* C(x)P(x*)T(x),C(x)P(x*)~T(x)}^{C(x)P(x*)}
v[2]: ∀x {3.0=* P(x*)T(x)~C(x),P(x*)~C(x)~T(x)}^{P(x*)~C(x)}
v[3]: ∃a {P(a*)T(a)}
v[4]: ∃a {C(a)}
c (Conclusion): ∃a {0,15.0=* C(a)}
test(verbose=False): Method used to test the example
e66i¶
description:
Example 66, p191, p225
Think of 100 people.
1. One of the disease psylicrapitis, and he is likely to be positive.
2. Of those who do not have the disease, 1 will also test positive.
How many of those who test positive do have the disease? Out of ?
v[0]: {1.0=* D()T(),1.0=* T()~D(),98.0=* ~D()}
v[1]: {D()T()}
c (Conclusion): {}
test(verbose=False): Method used to test the example
e66ii¶
description:
Example 66, p191, p225
Think of 100 people.
1. One of the disease psylicrapitis, and he is likely to be positive.
2. Of those who do not have the disease, 1 will also test positive.
How many of those who test positive do have the disease? Out of ?
v[0]: {1.0=* D()T(),1.0=* T()~D(),98.0=* ~D()}
v[1]: {T()}
c (Conclusion): {}
test(verbose=False): Method used to test the example
e67¶
description:
Example 67, p191, p220
Results of a recent survey of seventy-four chief executive officers indicate there
may be a link between childhood pet ownership and future career success. Fully 94%
of the CEOs, all of them employed within Fortune 500 companies, had possessed a dog,
a cat, or both, as youngsters.
v[0]: {94.0=* HadPet()IsCEO(),~IsCEO()}
v[1]: {HadPet()}
v[2]: {IsCEO()}
c (Conclusion): {0,94.0=* IsCEO()}
test(verbose=False): Method used to test the example
e69_part1¶
description:
Example 69, p192, p218
The suspect's DNA matches the crime sample.
If the suspect is not guilty, then the probability of such a DNA match is 1 in
a million
Is the suspect likely to be guilty?
v[0]: {Match(Suspect())}
v[1]: {0.000001=* Match(Suspect())~Guilty(Suspect()),~Guilty(Suspect())~Match(Suspect())}^{~Guilty(Suspect())}
c (Conclusion): {Guilty(Suspect())Match(Suspect()),0.000001=* Match(Suspect())~Guilty(Suspect())}
test(verbose=False): Method used to test the example
e69_part2¶
description:
The base class for all examples. It contains a series of views (v)
for operations and a conclusion (c).
v[0]: {Guilty(Suspect())Match(Suspect()),0.000001=* Match(Suspect())~Guilty(Suspect())}
v[1]: {999999.999999=* 0}^{Guilty(Suspect())Match(Suspect())}
v[2]: {Guilty(Suspect())}
v[0]: {999999.999999=* Guilty(Suspect())Match(Suspect()),0.000001=* Match(Suspect())~Guilty(Suspect())}
v[1]: {0,999999.999999=* Guilty(Suspect())}
test(verbose=False): Method used to test the example
e70¶
description:
Example 70, p194, p221
P1 Pat has either the disease or a benign condition
P2 If she has the disease, then she will have a certain symptom.
P3 In fact, she has the symptom
v[0]: {Benign(),Disease()}
v[1]: {Disease()Symptom()}^{Disease()}
v[2]: {Symptom()}
c (Conclusion): {Disease()Symptom()}
test(verbose=False): Method used to test the example
e71¶
description:
Examples 71 & 78, p209, p212
There is a box in which there is a yellow card or a brown card, but not both.
Given the preceding assertion, according to you, what is the probability of the following situation?
In the box there is a yellow card and there is not a brown card
v[0]: {B(brown())~B(yellow()),B(yellow())~B(brown())}
v[1]: {50.0=* 0}^{B(yellow())~B(brown())}
v[2]: {50.0=* 0}^{B(brown())~B(yellow())}
v[3]: {B(yellow())~B(brown())}
v[0]: {50.0=* B(brown())~B(yellow()),50.0=* B(yellow())~B(brown())}
v[1]: {0,50.0=* B(yellow())~B(brown())}
test(verbose=False): Method used to test the example
e72¶
description:
Example 72 & 80, p196, p213
There is a box in which there is at least a red marble or else there is a green
marble and there is a blue marble, but not all three marbles.
What is the probability of the following situation:
There is a red marble and a blue marble in the box?
v[0]: {B(b())B(g())~B(r()),B(r())~B(b()),B(r())~B(g())}
v[1]: {33.333333=* 0}^{B(b())B(g())~B(r())}
v[2]: {33.333333=* 0}^{B(r())~B(g())}
v[3]: {33.333333=* 0}^{B(r())~B(b())}
v[4]: {B(b())B(r())}
v[0]: {33.333333=* B(b())B(g())~B(r()),33.333333=* B(r())~B(b()),33.333333=* B(r())~B(g())}
v[1]: {0,33.333333=* B(b())B(r())}
test(verbose=False): Method used to test the example
e74¶
description:
Example 74, p197, p231
(includes two background commitments)
v[0]: {D(j())H(j()),H(j()),P(j())}
v[1]: {E(j()*)}
v[2]: ∀x {0.85=* D(x)E(x*),0.15=* E(x*)~D(x)}^{E(x*)}
v[3]: ∀x {0.1=* E(x*)H(x),0.9=* E(x*)~H(x)}^{E(x*)}
v[0]: {0.085=* D(j())E(j()*)H(j()),0.765=* D(j())E(j()*)~H(j()),0.015=* E(j()*)H(j())~D(j()),0.135=* E(j()*)~D(j())~H(j())}
v[1]: {D(j())H(j())}
test(verbose=False): Method used to test the example
e76¶
description:
Example 76 (guns and guitars), p199, p226, p229
(P1) The gun fired and the guitar was out of tune, or else someone was in the attic
(P1.5, see p228) Guns who triggers are pulled fire
(P2) The trigger (of the gun) was pulled. Does it follow that the guitar was out of
tune?
v[0]: {Attic(a()),Fired(i()*)Guitar(j())Gun(i())Outoftune(j())}
v[1]: ∀x {0,Fired(x*)Gun(x)Trigger(x)}^{Fired(x*)Gun(x)}
v[2]: {Trigger(i())}
c (Conclusion): {Fired(i()*)Guitar(j())Gun(i())Outoftune(j())Trigger(i())}
test(verbose=False): Method used to test the example
e81i¶
description:
Example 81, p213
There is a box in which there is a yellow card, or a brown card, but not both
Given the preceding assertion, according to you, what is the probability of the following situation?
In the box there is a yellow card
v[0]: {Box(Brown())~Box(Yellow()),Box(Yellow())~Box(Brown())}
c (Conclusion): {0,50.0=* Box(Yellow())}
prob (Probability): {Box(Yellow())}
test(verbose=False): Method used to test the example
e81ii¶
description:
Example 81, p213
There is a box in which there is a yellow card, or a brown card, but not both
Given the preceding assertion, according to you, what is the probability of the following situation?
In the box there is a yellow card and a brown card
v[0]: {Box(Brown())~Box(Yellow()),Box(Yellow())~Box(Brown())}
c (Conclusion): {0}
prob (Probability): {Box(Brown())Box(Yellow())}
test(verbose=False): Method used to test the example
e81iii¶
description:
Example 81, p213
There is a box in which there is a yellow card, or a brown card, but not both
Given the preceding assertion, according to you, what is the probability of the following situation?
In the box there is neither a yellow card nor a brown card
v[0]: {Box(Brown())~Box(Yellow()),Box(Yellow())~Box(Brown())}
c (Conclusion): {0}
prob (Probability): {~Box(Brown())~Box(Yellow())}
test(verbose=False): Method used to test the example
e82i¶
description:
Example 82, p213
There is a box in which if there is a yellow card then there is a brown card.
Given the preceding assertion, according to you, what is the probability of the
following situation?
In the box there is a yellow card.
v[0]: {Box(Brown())Box(Yellow())}^{Box(Yellow())}
c (Conclusion): {0,50.0=* Box(Yellow())}
prob (Probability): {Box(Yellow())}
test(verbose=False): Method used to test the example
e82ii¶
description:
Example 82, p213
There is a box in which if there is a yellow card then there is a brown card.
Given the preceding assertion, according to you, what is the probability of the
following situation?
In the box there is a yellow card and a brown card.
v[0]: {Box(Brown())Box(Yellow())}^{Box(Yellow())}
c (Conclusion): {0,50.0=* Box(Brown())Box(Yellow())}
prob (Probability): {Box(Brown())Box(Yellow())}
test(verbose=False): Method used to test the example
e82iii¶
description:
Example 82, p213
There is a box in which if there is a yellow card then there is a brown card.
Given the preceding assertion, according to you, what is the probability of the
following situation?
In the box there is a yellow card and there is not a brown card.
v[0]: {Box(Brown())Box(Yellow())}^{Box(Yellow())}
c (Conclusion): {0}
prob (Probability): {Box(Yellow())~Box(Brown())}
test(verbose=False): Method used to test the example
e82iv¶
description:
Example 82, p213
There is a box in which if there is a yellow card then there is a brown card.
Given the preceding assertion, according to you, what is the probability of the
following situation?
In the box there is neither a yellow card nor a brown card.
v[0]: {Box(Brown())Box(Yellow())}^{Box(Yellow())}
c (Conclusion): {0}
prob (Probability): {~Box(Brown())~Box(Yellow())}
test(verbose=False): Method used to test the example
e83i¶
description:
Example 83, p214
There is a box in which there is a red marble, or else there is a green
marble and there is a blue marble, but not all three marbles.
Given the preceding assertion, according to you, what is the probability of the
following situation?
There is a red marble and blue in marble in the box.
v[0]: {33.333333333333336=* Box(Blue())Box(Green()),33.333333333333336=* Box(Red()),33.333333333333336=* ~Box(Blue())~Box(Green())~Box(Red())}
c (Conclusion): {0}
prob (Probability): {Box(Blue())Box(Red())}
test(verbose=False): Method used to test the example
e83ii¶
description:
Example 83, p214
There is a box in which there is a red marble, or else there is a green
marble and there is a blue marble, but not all three marbles.
Given the preceding assertion, according to you, what is the probability of the
following situation?
There is a green marble and there is a blue marble.
v[0]: {33.333333333333336=* Box(Blue())Box(Green()),33.333333333333336=* Box(Red()),33.333333333333336=* ~Box(Blue())~Box(Green())~Box(Red())}
c (Conclusion): {0,33.333333333333336=* Box(Blue())Box(Green())}
prob (Probability): {Box(Blue())Box(Green())}
test(verbose=False): Method used to test the example
e84i¶
description:
Example 84, p215
There is a box in which there is a grey marble and either a white marble or
else a mauve marble but not all three marbles are in the box.
Given the preceding assertion, what is the probability of the following
situation?
In the box there is a grey marble and there is a mauve marble.
v[0]: {Box(Grey())Box(Mauve())~Box(White()),Box(Grey())Box(White())~Box(Mauve())}
c (Conclusion): {0,50.0=* Box(Grey())Box(Mauve())}
prob (Probability): {Box(Grey())Box(Mauve())}
test(verbose=False): Method used to test the example
e84ii¶
description:
Example 84, p215
There is a box in which there is a grey marble, or else a white marble, or else a mauve marble,
but no more than one marble.
Given the preceding assertion, what is the probability of the following
situation?
In the box there is a grey marble and there is a mauve marble.
v[0]: {Box(Grey())~Box(Mauve())~Box(White()),Box(Mauve())~Box(Grey())~Box(White()),Box(White())~Box(Grey())~Box(Mauve())}
c (Conclusion): {0}
prob (Probability): {Box(Grey())Box(Mauve())}
test(verbose=False): Method used to test the example
e85¶
description:
Example 85, p216
Easy partial probability inference
There is a box in which there is one and only one of these marbles: a
green marble, a blue marble, or a red marble. The probability that a green
marble is in the box is 0.6, and the probability that a blue marble is in
the box is 0.2.
What is the probability that a red marble is in the box?
v[0]: {Box(Blue()),Box(Green()),Box(Red())}
v[1]: {60.0=* Box(Green())}^{Box(Green())}
v[2]: {20.0=* Box(Blue())}^{Box(Blue())}
c (Conclusion): {0,20.0=* Box(Red())}
prob (Probability): {Box(Red())}
test(verbose=False): Method used to test the example
e86¶
description:
Example 86, p217
You have a hand of several cards with only limited information about it.
There is an ace and a queen or a king and a jack or a ten.
The probability that there is an ace and a queen is 0.6
The probability that there is a king and a jack is 0.2
What is the probability that there is a ten?
v[0]: {A()Q(),J()K(),X()}
v[1]: {60.0=* A()Q()}^{A()Q()}
v[2]: {20.0=* J()K()}^{J()K()}
c (Conclusion): {0,20.0=* X()}
prob (Probability): {X()}
test(verbose=False): Method used to test the example
e88¶
description:
Example 88, p233
P1: There is a 90% chance Superman can fly
P2: Clark is superman
C: There is a 90% chance Clark can fly
v[0]: {90.0=* CanFly(Superman())}
v[1]: {==(Clark(),Superman())}
v[2]: {==(Clark(),Superman()*)}
v[3]: {==(Clark(),Clark())}
c (Conclusion): {90.0=* CanFly(Clark())}
test(verbose=False): Method used to test the example
e90_condA¶
description:
Example 90, p249, p273
Imagine that you have been saving some extra money on the side to make some purchases,
and on your most recent visit to the video store you come across a special sale of a new
video. This video is one with your favourite actor or actress, and your favourite type of
movie (such as a comedy, drama, thriller etc.). This particular video that you are considering
is one you have been thinking about buying a long time. It is a available at a special sale price
of $14.99. What would you do in this situation?
v[0]: {do(Buy(Video()*)),~do(Buy(Video()*))}
c (Conclusion): {do(Buy(Video()*))}
v[0]: ∀x {Fun()}^{do(Buy(x*))}
v[0]: {1.0=+ 0}^{Fun()}
test(verbose=False): Method used to test the example
e90_condB¶
description:
Example 90, p249, p273
Imagine that you have been saving some extra money on the side to make some purchases,
and on your most recent visit to the video store you come across a special sale of a new
video. This video is one with your favourite actor or actress, and your favourite type of
movie (such as a comedy, drama, thriller etc.). This particular video that you are considering
is one you have been thinking about buying a long time. It is a available at a special sale price
of $14.99. What would you do in this situation?
v[0]: ∃a {do(Buy(Video()*)),do(Buy(a*))}
c (Conclusion): ∃a {do(Buy(Video()*)),do(Buy(a*))}
v[0]: ∀x {Fun()}^{do(Buy(x*))}
v[0]: {1.0=+ 0}^{Fun()}
test(verbose=False): Method used to test the example
e92_award¶
description:
Example 92, p253, p274
Imagine that you serve on the jury of an only-child sole-custody case following a relatively
messy divorce. The facts of the case are complicated by ambiguous economic, social, and
emotional considerations, and you decide to base your decision entirely on the following
few observations.
ParentA: average income, average health, average working hours, reasonable rapport with the
child, relatively social life.
ParentB: above-average income, very close relationship with the child, extremely active
social life, lots of work-related travel, minor health problems.
To which parent would you award sole custody of the child?
v[0]: {do(Award(ParentA()*)),do(Award(ParentB()*))}
c (Conclusion): {do(Award(ParentB()*))}
v[0]: ∀x {Custody(x*)}^{do(Award(x*))}
v[1]: ∀x {~Custody(x*)}^{do(Deny(x*))}
v[2]: {HighRapp(ParentB())LowTime(ParentB())MedRapp(ParentA())MedTime(ParentA())}
v[0]: ∀x {1.0=+ 0}^{Custody(x*)MedRapp(x)}
v[1]: ∀x {3.0=+ 0}^{Custody(x*)HighRapp(x)}
v[2]: ∀x {1.0=+ 0}^{Custody(x*)MedTime(x)}
v[3]: ∀x {1.0=+ 0}^{MedTime(x)~Custody(x*)}
v[4]: ∀x {2.0=+ 0}^{LowTime(x)~Custody(x*)}
test(verbose=False): Method used to test the example
e92_deny¶
description:
Example 92, p253, p274
Imagine that you serve on the jury of an only-child sole-custody case following a relatively
messy divorce. The facts of the case are complicated by ambiguous economic, social, and
emotional considerations, and you decide to base your decision entirely on the following
few observations.
ParentA: average income, average health, average working hours, reasonable rapport with the
child, relatively social life.
ParentB: above-average income, very close relationship with the child, extremely active
social life, lots of work-related travel, minor health problems.
To which parent would you deny sole custody of the child?
v[0]: {do(Deny(ParentA()*)),do(Deny(ParentB()*))}
c (Conclusion): {do(Deny(ParentB()*))}
v[0]: ∀x {Custody(x*)}^{do(Award(x*))}
v[1]: ∀x {~Custody(x*)}^{do(Deny(x*))}
v[2]: {HighRapp(ParentB())LowTime(ParentB())MedRapp(ParentA())MedTime(ParentA())}
v[0]: ∀x {1.0=+ 0}^{Custody(x*)MedRapp(x)}
v[1]: ∀x {3.0=+ 0}^{Custody(x*)HighRapp(x)}
v[2]: ∀x {1.0=+ 0}^{Custody(x*)MedTime(x)}
v[3]: ∀x {1.0=+ 0}^{MedTime(x)~Custody(x*)}
v[4]: ∀x {2.0=+ 0}^{LowTime(x)~Custody(x*)}
test(verbose=False): Method used to test the example
e93_grp1¶
description:
Example 93, p255, p276
The US is preparing for the outbreak of an unusual Asian disease, which
is expected to kill 600 people. There are two possible treatments (A) and (B)
with the following results:
(Group 1) (A) 400 people die. (B) Nobody dies with 1/3 chance, 600 people die with 2/3 chance.
Which treatment would you choose?
v[0]: {do(A()),do(B())}
c (Conclusion): {do(B())}
v[0]: {D(400.0*)}^{do(A())}
v[1]: {0.33=* D(0.0*),~D(0.0)}^{do(B())}
v[2]: {0.67=* D(600.0*),~D(600.0)}^{do(B())}
v[0]: ∀x {power(σ(1.0,log(σ(1.0,x))),-1.0)=+ 0}^{D(x*)}
v[1]: ∀x {σ(1.0,log(σ(1.0,x)))=+ 0}^{S(x*)}
test(verbose=False): Method used to test the example
new_e1¶
description:
The base class for all examples. It contains a series of views (v)
for operations and a conclusion (c).
v[0]: ∀x ∃a ∀y {P(x,a)Q(a,y)}
v[1]: ∃b ∀z {P(b,z)}
c (Conclusion): ∃b ∀x ∀z ∃a ∀y {P(b,z)P(x,a)Q(a,y)}
test(verbose=False): Method used to test the example
new_e2¶
description:
The base class for all examples. It contains a series of views (v)
for operations and a conclusion (c).
v[0]: ∃a ∀x {P(a)Q(x*)}
v[1]: ∀x ∃b {Q(x*)R(b)}^{Q(x*)}
c (Conclusion): ∃a ∀x ∃b {P(a)Q(x*)R(b)}
test(verbose=False): Method used to test the example
else_inquire¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: ∃a ∀x {P(a)Q(x*)}
v[1]: ∀x ∃b {Q(x*)R(b)}^{Q(x*)}
c (Conclusion): ∃a ∀x {P(a)Q(x*)}
test(verbose=False): Method used to test the example
else_merge¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: ∃a ∀x {P(a)Q(x*)}
v[1]: ∀x ∃b {Q(x*)R(b)}^{Q(x*)}
c (Conclusion): ∃a ∀x {P(a)Q(x*)}
test(verbose=False): Method used to test the example
else_suppose¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: ∃a ∀x {P(a)Q(x*)}
v[1]: ∀x ∃b {Q(x*)R(b)}^{Q(x*)}
c (Conclusion): ∃a ∀x {P(a)Q(x*)}
test(verbose=False): Method used to test the example
else_uni_prod¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: ∃a ∀x {P(a)Q(x*)}
v[1]: ∀x ∃b {Q(x*)R(b)}^{Q(x*)}
c (Conclusion): ∃a ∀x {P(a)Q(x*)}
test(verbose=False): Method used to test the example
else_query¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: ∃a ∀x {P(a)Q(x*)}
v[1]: ∀y ∃a {Q(y*)R(a)}^{Q(y*)}
c (Conclusion): ∃a ∀x {P(a)Q(x*)}
test(verbose=False): Method used to test the example
else_which¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: ∃a ∀x {P(a)Q(x*)}
v[1]: ∀y ∃a {Q(y*)R(a)}^{Q(y*)}
c (Conclusion): ∃a ∀x {P(a)Q(x*)}
test(verbose=False): Method used to test the example
new_e5¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: ∀x ∀y ∃a ∃b ∀z ∃c {P(a*)P(b)P(c)P(y)P(z)Q(x*)}
v[1]: ∃d ∃e ∃f {P(d*)Q(e*)Q(f*)}
c (Conclusion): ∃d ∃e ∃f {P(d*)Q(e*)Q(f*)}
test(verbose=False): Method used to test the example
new_e6_leibniz¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: ∃a ∃b {==(a,b)P(f(a),a)~P(f(b),a)}
v[1]: {}
c (Conclusion): {}
test(verbose=False): Method used to test the example
new_e7_aristotle¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: ∃a {~==(a,a)}
v[1]: {}
c (Conclusion): {}
test(verbose=False): Method used to test the example
new_e8¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: {t()=+ A()}
v[1]: {u()=* A()}
c (Conclusion): {u()=* t()=+ A()}
test(verbose=False): Method used to test the example
new_e9¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: ∀x {P(x*)}
v[1]: {P(j()*)}
c (Conclusion): {0}
test(verbose=False): Method used to test the example
new_e10¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: ∀x {f(x)=* A(x*)}
v[1]: ∃e {f(e)=* A(e*)}
c (Conclusion): ∃e {f(e)=* A(e*)}
test(verbose=False): Method used to test the example
new_e11¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: {f(12.0)=* A(12.0*)}
v[1]: ∃e {f(e)=* A(e*)}
c (Conclusion): ∃e {f(e)=* A(e*)}
test(verbose=False): Method used to test the example
new_e12¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: {A()}
v[1]: {}
c (Conclusion): {A()}
test(verbose=False): Method used to test the example
new_e13¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: {f(12.0)=* A(12.0*),B()}
c (Conclusion): {}
prob (Probability): ∃e {A(e*)}
test(verbose=False): Method used to test the example
new_e14¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: ∀x ∃y {A(f(x*))B(g(x*,y))}
v[1]: {A(f(j()*))}
c (Conclusion): ∃y {A(f(j()*))B(g(j()*,y))}
test(verbose=False): Method used to test the example
new_e15¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: ∃k {==(Clark(),Superman())Defeats(k,Superman())}
v[1]: {==(Clark()*,Superman())}
c (Conclusion): ∃k {==(Clark(),Clark())Defeats(k,Clark())}
test(verbose=False): Method used to test the example
new_e16¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: ∃k ∃x {==(Clark(),x)Defeats(k,x)}
v[1]: ∃x {==(Clark()*,x)}
c (Conclusion): ∃k {==(Clark(),Clark())Defeats(k,Clark())}
test(verbose=False): Method used to test the example
new_e17¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: ∃k ∃x {==(Clark(),x)do(Defeats(k,x))}
v[1]: ∃x {==(Clark()*,x)}
c (Conclusion): ∃k {==(Clark(),Clark())do(Defeats(k,Clark()))}
test(verbose=False): Method used to test the example
new_e18¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: {m()=* A()}
v[1]: {n()=* B()}
c (Conclusion): {m()**n()=* A()B()}
test(verbose=False): Method used to test the example
new_e19_first_atom_do_atom¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: ∃k {==(Clark(),Superman())Defeats(k,Superman())}
v[1]: {do(A())}
c (Conclusion): ∃k {==(Clark(),Superman())Defeats(k,Superman())}
test(verbose=False): Method used to test the example
new_e20_nested_issue_in_pred¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: ∃k {==(Clark(),Superman())Defeats(k,Superman())}
v[1]: {==(Clark(),f(Superman()*))}
c (Conclusion): ∃k {==(Clark(),Superman())Defeats(k,Superman())}
test(verbose=False): Method used to test the example
new_e21_supp_is_something¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: ∃k {==(Clark(),Superman())Defeats(k,Superman())}
v[1]: {==(Clark()*,Superman())}^{}
c (Conclusion): ∃k {==(Clark(),Superman())Defeats(k,Superman())}
test(verbose=False): Method used to test the example
new_e22_restrict_dep_rel_is_not_other¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: ∃k ∃x {==(Clark(),x)do(Defeats(k,x))}
v[1]: ∃y {==(Clark()*,y)}
c (Conclusion): ∃k ∃x {==(Clark(),x)do(Defeats(k,x))}
test(verbose=False): Method used to test the example
AnswerPotential¶
description:
The base class for all examples. It contains a series of views (v)
for operations and a conclusion (c).
v[0]: {1.0=* 2.0=+ A()B(),A()C(),0.4=* B()C()}
v[1]: {A()}
v[2]: {B()}
v[3]: {C()}
v[4]: {C()D()}
v[5]: {C()~B()}
c (Conclusion): {}
test(verbose=False): Method used to test the example
UniProduct¶
description:
The base class for all test types. Test types are mixins the define a test
type, but not the associated views.
v[0]: ∀x ∃a {E(x,a)P(x),~P(x*)}
v[1]: {P(j()*)}
c (Conclusion): ∃a {E(j(),a)P(j()),~P(j()*)}
test(verbose=False): Method used to test the example
QueryTest¶
description:
From page 173
v[0]: ∀x {S(j()*)S(m()*)T(x,j()),S(j()*)S(m()*)T(x,m())}
v[1]: ∀x ∃a {S(a*)T(x,a)}
c (Conclusion): ∀x ∃a {S(a*)T(x,a)}
test(verbose=False): Method used to test the example
QueryTest2¶
description:
From page 173
v[0]: ∀x {S(j()*)S(m()*)T(x,j()),S(j()*)S(m()*)T(x,m())}
v[1]: ∃a ∀x {S(a*)T(x,a)}
c (Conclusion): ∀x ∃a {S(a*)T(x,a)}
test(verbose=False): Method used to test the example