1. First Order Probabilistic Inference, by David Poole (2003)
Rodrigo de Salvo Braz

2. P(x1,x2,x3,x4) = P(x3|x1,x2)P(x1)P(x2|x4)P(x4)
Bayesian Networks x1 x2 x4 x3
P(x1,x2,x3,x4) = P(x3|x1,x2)P(x1)P(x2|x4)P(x4)

3. P(x1,x2,x3,x4)  1(x1)2(x1,x2, x3)3(x2,x4) 4(x4)
Factor Networks 3 x1 x2 x4 1 4 2 x3
P(x1,x2,x3,x4)  1(x1)2(x1,x2, x3)3(x2,x4) 4(x4)

4. Variable Elimination (VE) in Factor Networksz1 z1 1 5 4 2 z4 z2 z4 y z2 3 z3 z3
...y1(z1,y)2(z2,y)3(z3,y)4(z4,y)...
...1(z1,y)2(z2,y)3(z3,y)4(z4,y)...
...5 (z1,z2,z3,z4)...

5. VE in Factor Networks 3 x1 x2 x4 1 4 2 x3
Query P(x2) = ?

6. VE in Factor Networks 5 x1 x2 1 2 x3
Query P(x2) = ?

7. VE in Factor Networks 5 x2 6 x3
Query P(x2) = ?

8. VE in Factor Networks 5 x2 7
P(x2)  5(x2)7 (x2)

9. First Order Probabilistic Inference
If a person P has been in a country C and C speaks a language L, then P speaks L with probability 0.4 else P speaks L with probability 0.01.

10. First Order Probabilistic Inference
If a person P has been in a country C and C speaks a language L, then P speaks L with probability 0.4 else P speaks L with probability 0.01.
h_b
cn_spks
spks  T
0.4 F
0.6
0.01
0.5
h_b (P,C)
cn_spks(C,L) 
spks(P,L)

11. First Order Probabilistic Inference
domain: nutland, crazynia, john, mary, french P(john has been to nutland | john speaks french, mary has been to nutland, mary speaks french) = ?

12. First Order Probabilistic Inference
domain: nutland, crazynia, john, mary, french P(john has been to nutland | john speaks french, mary has been to nutland, mary speaks french) = ?
h_b(john,nutland)
cn_spks(nutland,french)
h_b(mary,nutland)  
spks(john,french)
spks(mary,french)  
h_b(john,france)
cn_spks(crazynia,french)
h_b(mary,france)

13. First Order Probabilistic Inference
Full grounding unfeasible, size increases fast.

14. First Order Probabilistic Inference
Full grounding unfeasible, size increases fast.
We may know that there are, e.g., 200 countries, 300 languages, without knowing any difference between them.
...
...
h_b(john,c1)
spks(john,l1) 
cn_spks(c1,l1)
h_b(john,c100)
cn_spks(c100,l1) 
spks(john,l1)
...
...
...
...
h_b(john,c1)
spks(john,l200) 
cn_spks(c1,l200)
h_b(john,c100)
cn_spks(c100,l200) 
spks(john,l200)

15. First Order Probabilistic Inference
p(X)
Domain U = {a,b,...,z,a1,...} p(a), p(b), ..., q(a,b), ... are random variables. X, Y are logical variables. Represents infinite factors: 
q(X,Y)
p(a)
p(b)
p(z)
p(a)
p(b)
p(z)
p(a)
p(b)
p(z)    
p(a) 
p(b) 
p(z)
...
... 
p(a) 
p(b) 
p(z)
q(a,a)
q(b,a)
q(z,a)
. . .
. . .
. . .
q(a,b)
q(b,b)
q(z,b)   
q(a,c)
q(b,c)
q(z,c)   
q(a,z)
. . .
q(b,z)
. . .
q(z,z)
. . .
q(a,a1)
q(b,a1)
q(z,a1)

16. Implicit representation
Keep it implicit: the parameterized factor (parfactor) 
p(a)
q(a,a1)
q(a,z)
q(a,c)
q(a,b)
q(a,a)
. . .
...
p(b)
q(b,a1)
q(b,z)
q(b,c)
q(b,b)
q(b,a)
p(z)
q(z,a1)
q(z,z)
q(z,c)
q(z,b)
q(z,a)
p(X)  
q(X,Y)
{X  U, Y  U}

17. Implicit representation With Constraints
p(a)
p(X)
p(a)
p(a)
p(a)      
p(a) 
p(a)
q(a,a)
q(a,b)
. . . 
q(a,Y)
q(X,Y)
q(a,c) 
q(a,z)
. . .
q(a,a1)
{Y  U}
{X = a, Y  U}

18. Variable Elimination (VE) Example
t(W) 1
{W  b}
p(W) 3 2
q(X,Y)
p(Y)
p(V)
s(V) {}
{V  c}
We want to eliminate variables with predicate p

19. VE Example – Grounding ... ... ... ... ... ... {W  b}
t(a)
t(b)
t(c)
t(d)
...
{W  b} 1 1 1 1
...
p(a)
p(b)
p(c)
p(d) 3
q(a,a) 2
p(a) 3
p(a)
s(a)
q(b,a) 2
What are the common variables on p?
p(b)
s(b)
... 2 3
q(a,b)
p(c)
s(c)
p(b) 2 3
s(d)
q(b,b)
p(d)
...
...
...
{V  c} {}

20. VE Example – Grounding ... ... ... ... ... ... {W  b}
t(a)
t(b)
t(c)
t(d)
...
{W  b} 1 1 1 1
...
p(a)
p(b)
p(c)
p(d) 3
q(a,a) 2
p(a) 3
p(a)
s(a)
q(b,a) 2
Common variables: p(Y’){Y’b,Y’c}
p(b)
s(b)
... 2 3
q(a,b)
p(c)
s(c)
p(b) 2 3
s(d)
q(b,b)
p(d)
...
...
...
{V  c} {}

21. VE Example – Common variables
t(W) 1
{W  b}
p(W) 3 2
q(X,Y)
p(Y)
p(V)
s(V) {}
{V  c}
Unification(p(Y), p(W){W  b}, p(V){V  c}) = p(Y’){Y=V=W=Y’,Y’  b, Y’  c}

22. VE Example – Unified parfactors only
t(Y’) 1
{Y’b,Y’c}
p(Y’) 3 2
q(X,Y’)
p(Y’)
p(Y’)
s(Y’)
{Y’b,Y’c}
{Y’b,Y’c}
What about the remaining bindings?

23. VE Example – Splitting parfactors
Unification
Residual
t(W)
t(W)
t(W) - = 1 1 1
{Wb,Wc}
{W = c}
{W  b}
p(W)
p(W)
p(W) W W W - = a b c
... a b c
... a b c
...

24. VE Example – Splitting parfactors
Unification
Residual 2 2 2 - =
p(V)
s(V)
p(V)
s(V)
p(V)
s(V)
{V  c}
{Vb,Vc}
{V = b} V V V - = a b c
... a b c
... a b c
...

25. VE Example – Splitting parfactors
Unification
Residual 3 3 3 - =
q(X,Y)
p(Y)
q(X,Y)
p(Y)
q(X,Y)
p(Y) {}
{Yb,Yc}
{Y{b,c}} Y Y Y - = a b c
... a b c
... a b c
...

26. VE Example – Grounded view
for each Y
t(a) 1
q(a,a) 3
p(a)
s(a)
q(b,a) 2 3
...
t(d) 1
q(a,d) 3
p(d)
s(d)
q(b,d) 2 3
...

27. VE Example – Grounded view
t(a)
for each Y
for each X,Y
11/|X|
t(a)
q(a,a)
p(a)
s(a) 1 3
21/|X|
q(a,a) 3
p(a)
s(a)
t(a)
q(b,a) 2
11/|X| 3
...
q(b,a)
p(a)
s(a) 3
...
21/|X|
t(d)
11/|X|
t(d)
q(a,d)
p(d)
s(d) 1 3
21/|X|
q(a,d) 3
p(d)
s(d)
t(d)
q(b,d) 2
11/|X| 3
...
q(b,d)
p(d)
s(d) 3
...
21/|X|

28. VE Example – Grounded view
t(a)
for each Y
for each X,Y
11/|X|
t(a)
t(a)
q(a,a)
p(a)
s(a)
q(a,a)
s(a) 1 3
21/|X|
q(a,a)
4 =p(Y) 311/|X|21/|X| 3
p(a)
s(a)
t(a)
q(b,a) 2
t(a)
11/|X| 3
...
q(b,a)
p(a)
s(a)
q(b,a)
s(a)
... 3
...
21/|X| 4
t(d)
t(d)
11/|X|
t(d)
q(a,d)
p(d)
s(d)
q(a,d)
s(d) 3
21/|X| 4 1
q(a,d) 3
p(d)
s(d)
t(d)
q(b,d) 2
11/|X|
t(d) 3
...
q(b,d)
p(d)
s(d)
q(b,d)
s(d)
... 3
...
21/|X| 4

29. VE Example – Grounded view
t(a)
for each Y
for each X,Y
11/|X|
t(a)
t(a)
q(a,a)
p(a)
s(a)
q(a,a)
s(a) 1 3
21/|X| 4
q(a,a) 3
p(a)
s(a)
t(a)
q(b,a) 2
t(a)
11/|X| 3
t(Y’)
...
q(b,a)
p(a)
s(a)
q(b,a)
s(a) 3
...
21/|X|
... 4
q(X,Y’)
s(Y’) 4
t(d)
{Y’b,Y’c}
t(d)
Emphasize that this is almost what intuition says, but not quite.
11/|X|
t(d)
q(a,d)
p(d)
s(d)
q(a,d)
s(d) 1 3
21/|X| 4
q(a,d) 3
p(d)
s(d)
t(d)
q(b,d) 2
11/|X|
t(d) 3
...
q(b,d)
p(d)
s(d)
q(b,d)
s(d)
... 3
...
21/|X|
... 4

30. VE Example – Lifted view
t(Y’) 1
{Y’b,Y’c}
p(Y’) 3 2
q(X,Y’)
p(Y’)
p(Y’)
s(Y’)
{Y’b,Y’c}
{Y’b,Y’c}

31. VE Example – Lifted view
t(Y’) 1
{Y’b,Y’c}
p(Y’) 3 2
q(X,Y’)
p(Y’)
p(Y’)
s(Y’)
{Y’b,Y’c}
{Y’b,Y’c}
t(Y’)
11/|X|
p(Y’)
s(Y’)
21/|X| 3
q(X,Y’)
{Y’b,Y’c}
{Y’b,Y’c}
{Y’b,Y’c}

32. VE Example – Lifted view
t(Y’) 1
{Y’b,Y’c}
p(Y’) 3 2
t(Y’)
q(X,Y’)
p(Y’)
p(Y’)
s(Y’)
q(X,Y’)
s(Y’)
{Y’b,Y’c}
{Y’b,Y’c}
p(Y) 311/|X|21/|X|
{Y’b,Y’c}
t(Y’)
11/|X|
p(Y’)
s(Y’)
21/|X| 3
q(X,Y’)
{Y’b,Y’c}
{Y’b,Y’c}
{Y’b,Y’c}

33. VE Example 2 – Losing logic variables1 2
q(X)
p(Y)
p(Y)
s(Z) {} {}
No constraints, so already unified;
Logical variable Y will disappear.

34. VE Example 2 – Grounded view
for each Y 1 2
q(a)
s(a)
p(a) 2 1
q(b)
s(b)
...
... 1 2
q(a)
s(a)
p(b) 2 1
q(b)
s(b)
...
...

35. VE Example 2 – Grounded view
for each Y
for each X,Y,Z
11/|Z|
21/|X|
q(a)
p(a)
s(a)
11/|Z|
21/|X| 1 2
q(a)
p(a)
s(b)
q(a)
s(a)
p(a)
... 2 1
11/|Z|
21/|X|
q(b)
s(b)
q(b)
p(a)
s(a)
...
...
11/|Z|
21/|X|
q(b)
p(a)
s(b)
...
11/|Z|
21/|X|
q(a)
p(b)
s(a)
11/|Z|
21/|X| 1 2
q(a)
p(b)
s(b)
q(a)
s(a)
...
p(b) 2 1
11/|Z|
21/|X|
q(b)
s(b)
q(b)
p(b)
s(a)
11/|Z|
21/|X|
...
...
q(b)
p(b)
s(b)
...

36. VE Example 2 – Grounded view
for each Y
for each X,Y,Z
=p(Y) 11/|Z|21/|X|
11/|Z|
21/|X| 
q(a)
p(a)
s(a)
q(a)
s(a)
11/|Z|
21/|X| 1 2 
q(a)
p(a)
s(b)
q(a)
s(a)
q(a)
s(b)
p(a)
... 2
... 1
11/|Z|
21/|X| 
q(b)
s(b)
q(b)
p(a)
s(a)
q(b)
s(a)
...
...
11/|Z|
21/|X| 
q(b)
p(a)
s(b)
q(b)
s(b)
...
...
11/|Z|
21/|X| 
q(a)
p(b)
s(a)
q(a)
s(a)
11/|Z|
21/|X|  1 2
q(a)
p(b)
s(b)
q(a)
s(b)
q(a)
s(a)
...
...
p(b) 2 1 
11/|Z|
21/|X|
q(b)
s(b)
p(b)
q(b)
s(a)
q(b)
s(a) 
11/|Z|
21/|X|
...
...
p(b)
s(b)
q(b)
s(b)
q(b)
...
...

37. VE Example 2 – Grounded view
for each Y
for each X,Y,Z
=p(Y) 11/|Z|21/|X|
11/|Z|
21/|X| 
q(a)
p(a)
s(a)
q(a)
s(a)
11/|Z|
21/|X| 1 2 
q(a)
p(a)
s(b)
q(a)
s(a)
q(a)
s(b)
p(a)
... 2
... 1
11/|Z|
21/|X| 
q(b)
s(b)
q(b)
p(a)
s(a)
q(b)
s(a)
...
...
11/|Z|
21/|X| 
q(b)
p(a)
s(b)
q(b)
s(b)
|Y| repetitions each
...
...
11/|Z|
21/|X| 
q(a)
p(b)
s(a)
q(a)
s(a)
11/|Z|
21/|X|  1 2
q(a)
p(b)
s(b)
q(a)
s(b)
q(a)
s(a)
...
...
p(b) 2 1 
11/|Z|
21/|X|
q(b)
s(b)
p(b)
q(b)
s(a)
q(b)
s(a) 
11/|Z|
21/|X|
...
...
q(b)
p(b)
s(b)
q(b)
s(b)
...
...

38. VE Example 2 – Grounded view
for each Y
for each X,Y,Z
=p(Y) 11/|Z|21/|X|
11/|Z|
21/|X| 
q(a)
p(a)
s(a)
q(a)
s(a)
11/|Z|
21/|X| 1 2 
q(a)
p(a)
s(b)
q(a)
s(a)
q(a)
s(b)
p(a)
... 2
... 1
11/|Z|
21/|X| 
|Y|
q(b)
s(b)
q(b)
p(a)
s(a)
q(b)
s(a)
q(a)
s(a)
...
...
11/|Z|
21/|X| 
|Y|
q(b)
p(a)
s(b)
q(b)
s(b)
q(a)
s(b)
...
...
...
|Y|
11/|Z|
21/|X| 
q(b)
s(a)
q(a)
p(b)
s(a)
q(a)
s(a)
|Y|
11/|Z|
21/|X|  1
q(b)
s(b) 2
q(a)
p(b)
s(b)
q(a)
s(b)
q(a)
s(a)
...
...
...
p(b) 2 1 
11/|Z|
21/|X|
q(b)
s(b)
q(b)
s(a)
q(b)
p(b)
s(a) 
11/|Z|
21/|X|
...
...
p(b)
s(b)
q(b)
s(b)
q(b)
...
...

39. VE Example 2 – Grounded view
for each X,Y,Z
=p(Y) 11/|Z|21/|X|
11/|Z|
21/|X| 
q(a)
p(a)
s(a)
q(a)
s(a)
11/|Z|
21/|X| 
q(a)
p(a)
s(b)
q(a)
s(b)
...
...
11/|Z|
21/|X| 
|Y|
q(b)
p(a)
s(a)
q(b)
s(a)
q(a)
s(a)
11/|Z|
21/|X| 
|Y|
q(b)
p(a)
s(b)
q(b)
s(b)
q(a)
s(b)
|Y|
...
...
...
q(X)
s(Z)
|Y|
11/|Z|
21/|X| 
q(b)
s(a) {}
q(a)
p(b)
s(a)
q(a)
s(a)
|Y|
11/|Z|
21/|X| 
q(b)
s(b)
q(a)
p(b)
s(b)
q(a)
s(b)
...
...
... 
11/|Z|
21/|X|
q(b)
p(b)
q(b)
s(a)
s(a) 
11/|Z|
21/|X|
q(b)
p(b)
s(b)
q(b)
s(b)
...
...

40. VE Example 2 – Lifted view1 2
q(X)
p(Y)
p(Y)
s(Z) {} {}
(p(Y) 11/|Z|21/|X|)|Y|
q(X)
s(Z) {}

41. General Case ... q1(X1,Z) q2(X2,Z) qn(Xn,Z) 1 2 n p(Z,Y) p(Z,Y)

42. (q1(X1,Z),…,qn(Xn,Z)) = p(Z,Y)ii(Xi,Z,Y)|C(Y)|/|C(X\Xi)|
General Case
q1(X1,Z)
q2(X2,Z)
qn(Xn,Z)
... 1 2 n
p(Z,Y)
p(Z,Y)
p(Z,Y)
{C}
{C}
{C} 
{C}
...
q1(X1,Z)
q2(X2,Z)
qn(X1,Z)
(q1(X1,Z),…,qn(Xn,Z)) = p(Z,Y)ii(Xi,Z,Y)|C(Y)|/|C(X\Xi)|

43. Conclusions Clear semantics;
Probabilistic first order without grounding;
Specialize parfactors only as needed through splitting them;
Problem: no functions, but this can be fixed.

44. The End

45. VE Example 2 – Grounded view Master
for each Y
for each X,Y,Z
=p(Y) 11/|Z|21/|X|
11/|Z|
21/|X| 
q(a)
p(a)
s(a)
q(a)
s(a)
11/|Z|
21/|X| 1 2 
q(a)
p(a)
s(b)
q(a)
s(a)
q(a)
s(b)
p(a)
... 2
... 1
11/|Z|
21/|X| 
|Y|
q(b)
s(b)
q(b)
p(a)
s(a)
q(b)
s(a)
q(a)
s(a)
...
...
11/|Z|
21/|X| 
|Y|
q(b)
p(a)
s(b)
q(b)
s(b)
q(a)
|Y| repetitions each
s(b)
|Y|
...
...
...
q(X)
s(Z)
|Y|
11/|Z|
21/|X| 
q(b)
s(a) {}
q(a)
p(b)
s(a)
q(a)
s(a)
|Y|
11/|Z|
21/|X|  1
q(b)
s(b) 2
q(a)
p(b)
s(b)
q(a)
s(b)
q(a)
s(a)
...
...
...
p(b) 2 1 
11/|Z|
21/|X|
q(b)
s(b)
q(b)
s(a)
q(b)
p(b)
s(a) 
11/|Z|
21/|X|
...
...
p(b)
s(b)
q(b)
s(b)
q(b)
...
...

46. (q1(X1,Z),…,qn(Xn,Z)){C} = p(Z,Y)ii(Xi,Z,Y)|C(Y)|/|C(X – Xi)|
General Case
Parfactors i(qi(Xi,Z),p(Z,Y)){C}, i = 1,…,n
Same set of constraints C for all of them
Xi: logic variables unique to i
p(Z,Y): literal being eliminated
qi(Xi,Z): combination of all other literals in i
Y: logic variables unique to p
Z: logic variables in both p and all qi’s
X: sequence X1,…,Xn
(q1(X1,Z),…,qn(Xn,Z)){C} = p(Z,Y)ii(Xi,Z,Y)|C(Y)|/|C(X – Xi)|

47. Variable Elimination Same Bindings
p(X)
r(Y)
p(X) 2
q(X,Y) 
q(X,Y) 2
{X  U, Y  U}
q(X,Y)
r(Y)
{X  U, Y  U}
{X  U, Y  U}
X  U
X  U
X  U
Y  U
Y  U
Y  U

48. Variable Elimination Different Bindings, Same Variables
p(X)
r(Y) 2
q(X,Y) 
{X = a, Y  U}
No one to one correspondence!
q(X,Y)
{X  U, Y  U}
X  U
X = a
Y  U
Y  U

49. Splitting  p(X) p(X) p(X)    q(X,Y) q(X,Y) q(X,Y)
{X  U, Y  U}
{X = a, Y  U}
{X  a, Y  U}
X  U
X = a
X  a
Y  U
Y  U
Y  U

50. Variable Elimination After Splitting
p(X)
r(Y)
p(X) 2
q(X,Y) 
q(X,Y) 2
{X = a, Y  U}
q(X,Y)
r(Y)
{X = a, Y  U}
{X = a, Y  U}
X = a
X = a
X = a
Y  U
Y  U
Y  U

51. Example - Grounding 3
t(a)
t(b)
t(c)
t(d)
q(a,a)
p(a)
... 3 1 1 1 1
q(b,a)
p(a)
...
...
p(a)
p(b)
p(c)
p(d) 3
q(a,b)
p(b) 2
s(a) 3
p(a)
Variables in common: p(Y’){Y’b,Y’c}
q(b,b)
p(b) 2
p(b)
s(b)
... 3 2
q(a,c)
p(c)
p(c)
s(c) 3 2
q(b,c)
p(c)
p(d)
s(d)
... 3
...
q(a,d)
p(d)
...

52. Probabilistic Propositional Logic
0.2: q
0.8: p  q
0.3: q  r
1: (p  q)  r  
p  q
0.2 q p  p  

53. VE Example – Lifted view
t(Y’) 1
{Y’b,Y’c}
p(Y’) 3 2
q(X,Y’)
p(Y’)
p(Y’)
s(Y’)
{Y’b,Y’c}
{Y’b,Y’c}
t(d) 1
q(a,d) 3
p(d)
s(d)
q(b,d) 2
... 3

54. VE Example – Lifted view
t(Y’)
{Y’b,Y’c} 1
{Y’b,Y’c} 2
p(Y’)
t(Y’)
p(Y’)
s(Y’)
11/|X|
{Y’b,Y’c} 3
q(X,Y’)
p(Y’)
p(Y’)
p(Y’)
s(Y’)
q(X,Y’)
p(Y’) 3
21/|X|
{Y’b,Y’c}
{Y’b,Y’c}
{Y’b,Y’c}
t(d)
11/|X|
t(d)
q(a,d)
p(d)
s(d) 1 3
21/|X|
q(a,d) 3
p(d)
s(d)
t(d)
q(b,d) 2
11/|X|
... 3
q(b,d)
p(d)
s(d) 3
...
21/|X|

55. VE Example – Lifted view
t(Y’)
{Y’b,Y’c} 1
{Y’b,Y’c} 2
p(Y’)
t(Y’)
p(Y’)
s(Y’)
t(Y’)
11/|X|
s(Y’)
{Y’b,Y’c}
q(X,Y’) 3
q(X,Y’)
p(Y’)
p(Y’)
p(Y’)
s(Y’) 4
q(X,Y’)
p(Y’) 3
21/|X|
{Y’b,Y’c}
{Y’b,Y’c}
{Y’b,Y’c}
{Y’b,Y’c}
t(d)
t(d)
11/|X|
t(d)
q(a,d)
p(d)
s(d)
q(a,d)
s(d) 1 3
21/|X| 4
q(a,d) 3
p(d)
s(d)
t(d)
q(b,d) 2
11/|X|
t(d)
... 3
q(b,d)
p(d)
s(d)
q(b,d)
s(d) 3
...
21/|X| 4

56. VE Example – Lifted view
t(Y’)
{Y’b,Y’c} 1
{Y’b,Y’c} 2
p(Y’)
t(Y’)
p(Y’)
s(Y’)
t(Y’)
11/|X|
q(X,Y’)
s(Y’)
{Y’b,Y’c} 3
q(X,Y’)
p(Y’)
p(Y’)
p(Y’)
s(Y’) 4
q(X,Y’)
t(Y’)
p(Y’) 3
21/|X|
{Y’b,Y’c}
q(X,Y’)
s(Y’)
{Y’b,Y’c}
{Y’b,Y’c}
{Y’b,Y’c} 4
t(d)
{Y’b,Y’c}
t(d)
11/|X|
t(d)
q(a,d)
p(d)
s(d)
q(a,d)
s(d) 4 1 3
21/|X|
q(a,d) 3
p(d)
s(d)
t(d)
q(b,d) 2
11/|X|
t(d)
... 3
q(b,d)
p(d)
s(d)
q(b,d)
s(d) 3
...
21/|X| 4

57. Probabilistic Propositional Logic
0.2: q
0.8: p  q
0.3: q  r
1: (p  q)  r

58. Probabilistic Propositional Logic
0.3 
0.2 q
0.2: q
0.8: p  q
0.3: q  r
1: (p  q)  r
p  q
0.8  p p 

59. Probabilistic Propositional Logic
0.2: q
0.8: p  q
0.3: q  r
1: (p  q)  r

60. Probabilistic Propositional Logic
0.3 
0.2 q
0.2: q
0.8: p  q
0.3: q  r
1: (p  q)  r
p  q
0.8  p p 

61. Probabilistic Propositional Logic
0.2: q
0.8: p  q
0.3: q  r
1: (p  q)  r

62. Probabilistic Propositional Logic
0.4 if t, 0.5 otherwise
noisy-or of (3) and (4)
1 if p  s and q 1 if p  s and q 0.3 if q and (p  s) 0.5 if q and (p  s) t r
0.2 q
(1) 0.2: q
(2) 0.8: p  q
(3) 0.3: q  r
(4) 1: (p  s)  r
(5) 0.4: t  r
0.8 if p, 0.5 otherwise p s