1. Machine Learning: Symbol-based9
Machine Learning: Symbol-based
9.0 Introduction
9.1 A Framework for Symbol-based Learning
9.2 Version Space Search
9.3 The ID3 Decision Tree Induction Algorithm
9.4 Inductive Bias and Learnability
9.5 Knowledge and Learning
9.6 Unsupervised Learning
9.7 Reinforcement Learning
9.8 Epilogue and References
9.9 Exercises
Additional source used in preparing the slides:
Jean-Claude Latombe’s CS121 (Introduction to Artificial Intelligence) lecture notes, (version spaces, decision trees)
Tom Mitchell’s machine learning notes (explanation based learning)

2. Chapter Objectives
Learn about several “paradigms” of symbol-based learning
Learn about the issues in implementing and using learning algorithms
The agent model: can learn, i.e., can use prior experience to perform better in the future

3. A learning agent Critic KB environment Learning element sensors
actuators

4. A general model of the learning process

5. A learning game with playing cards
I would like to show what a full house is. I give you examples which are/are not full houses:
6 6 6 9 9 is a full house
6 6 6 6 9 is not a full house
3 3 3 6 6 is a full house
1 1 1 6 6 is a full house
Q Q Q 6 6 is a full house
1 2 3 4 5 is not a full house
1 1 3 4 5 is not a full house
1 1 1 4 5 is not a full house
1 1 1 4 4 is a full house

6. A learning game with playing cards
If you haven’t guessed already, a full house is three of a kind and a pair of another kind.
6 6 6 9 9 is a full house
6 6 6 6 9 is not a full house
3 3 3 6 6 is a full house
1 1 1 6 6 is a full house
Q Q Q 6 6 is a full house
1 2 3 4 5 is not a full house
1 1 3 4 5 is not a full house
1 1 1 4 5 is not a full house
1 1 1 4 4 is a full house

7. Intuitively,
I’m asking you to describe a set. This set is the concept I want you to learn.
This is called inductive learning, i.e., learning a generalization from a set of examples.
Concept learning is a typical inductive learning problem: given examples of some concept, such as “cat,” “soybean disease,” or “good stock investment,” we attempt to infer a definition that will allow the learner to correctly recognize future instances of that concept.

8. Supervised learning
This is called supervised learning because we assume that there is a teacher who classified the training data: the learner is told whether an instance is a positive or negative example of a target concept.

9. Supervised learning?
This definition might seem counter intuitive. If the teacher knows the concept, why doesn’t s/he tell us directly and save us all the work?
The teacher only knows the classification, the learner has to find out what the classification is. Imagine an online store: there is a lot of data concerning whether a customer returns to the store. The information is there in terms of attributes and whether they come back or not. However, it is up to the learning system to characterize the concept, e.g,
If a customer bought more than 4 books, s/he will return.
If a customer spent more than $50, s/he will return.

10. Rewarded card example
Deck of cards, with each card designated by [r,s], its rank and suit, and some cards “rewarded”
Background knowledge in the KB: ((r=1)  …  (r=10))  NUM (r) ((r=J)  (r=Q)  (r=K))  FACE (r) ((s=S)  (s=C))  BLACK (s) ((s=D)  (s=H))  RED (s)
Training set: REWARD([4,C])  REWARD([7,C])  REWARD([2,S])  REWARD([5,H])  REWARD([J,S])

11. Rewarded card example
Training set: REWARD([4,C])  REWARD([7,C])  REWARD([2,S])  REWARD([5,H])  REWARD([J,S])
Card In the target set?
4 yes
7 yes
2 yes
5 no
J no
Possible inductive hypothesis, h,:
h = (NUM (r)  BLACK (s)  REWARD([r,s])

12. Learning a predicate
Set E of objects (e.g., cards, drinking cups, writing instruments)
Goal predicate CONCEPT (X), where X is an object in E, that takes the value True or False (e.g., REWARD, MUG, PENCIL, BALL)
Observable predicates A(X), B(X), … (e.g., NUM, RED, HAS-HANDLE, HAS-ERASER)
Training set: values of CONCEPT for some combinations of values of the observable predicates
Find a representation of CONCEPT of the form CONCEPT(X)  A(X)  ( B(X) C(X) )

13. How can we do this?
Go with the most general hypothesis possible: “any card is a rewarded card” This will cover all the positive examples, but will not be able to eliminate any negative examples.
Go with the most specific hypothesis possible: “the rewarded cards are 4, 7, 2” This will correctly sort all the examples in the training set, but it is overly specific, will not be sort any new examples.
But the above two are good starting points.

14. Version space algorithm
What we want to do is start with the most general and specific hypotheses, and when we see a positive example, we minimally generalize the most specific hypotheses when we see a negative example, we minimally specialize the most general hypothesis
When the most general hypothesis and the most specific hypothesis are the same, the algorithm has converged, this is the target concept

15. potential target concepts
Pictorially + - - - + ? ? + ? - - + - + ? ? - + - - + ? ? + ? + + - + +
boundary of G - - - - - - - - - - ? - ? + + - - - + - + + ? + + + + ? - ? - - - - -
boundary of S
potential target concepts

16. Hypothesis space
When we shrink G, or enlarge S, we are essentially conducting a search in the hypothesis space
A hypothesis is any sentence h of the form CONCEPT(X)  A(X)  ( B(X) C(X) ) where, the right hand side is built with observable predicates
The set of all hypotheses is called the hypothesis space, or H
A hypothesis h agrees with an example if it gives the correct value of CONCEPT

17. Size of the hypothesis space
n observable predicates
2^n entries in the truth table
A hypothesis is any subset of observable predicates with the associated truth tables: so there are 2^(2^n) hypotheses to choose from: BIG!
n=6  2 ^ 64 = 1.8 x 10 ^ BIG!
Generate-and-test won’t work. 2 2n

18. Simplified Representation for the card problem
For simplicity, we represent a concept by rs, with:
r = a, n, f, 1, …, 10, j, q, k
s = a, b, r, , , ,  For example:
n represents: NUM(r)  (s=)  REWARD([r,s])
aa represents:
ANY-RANK(r)  ANY-SUIT(s)  REWARD([r,s])

19. Extension of an hypothesis
The extension of an hypothesis h is the set of objects that verifies h.
For instance,
the extension of f is: {j, q, k}, and
the extension of aa is the set of all cards.

20. More general/specific relation
Let h1 and h2 be two hypotheses in H
h1 is more general than h2 iff the extension of h1 is a proper superset of the extension of h2
For instance,
aa is more general than f,
f is more general than q,
fr and nr are not comparable

21. More general/specific relation (cont’d)
The inverse of the “more general” relation is the “more specific” relation
The “more general” relation defines a partial ordering on the hypotheses in H

22. A subset of the partial order for cardsaa na ab nb n 4 4b a 4a

23. G-Boundary / S-Boundary of V
An hypothesis in V is most general iff no hypothesis in V is more general
G-boundary G of V: Set of most general hypotheses in V
An hypothesis in V is most specific iff no hypothesis in V is more general
S-boundary S of V: Set of most specific hypotheses in V

24. Example: The starting hypothesis spaceaa na ab nb n 4 4b a 4a aa 4 1 k … S

25. 4 is a positive example
We replace every hypothesis in S whose extension does not contain 4 by its generalization set aa na ab
The generalization set of a hypothesis h is the set of the hypotheses that are immediately more general than h 4a nb a 4b n 4
Generalization
set of 4

26. 7 is the next positive example
Minimally generalize the most specific hypothesis set aa
We replace every hypothesis in S whose extension does not contain 7 by its generalization set na ab 4a nb a 4b n 4

27. 7 is positive(cont’d)
Minimally generalize the most specific hypothesis set aa na ab 4a nb a 4b n 4

28. 7 is positive (cont’d)
Minimally generalize the most specific hypothesis set aa na ab 4a nb a 4b n 4

29. 5 is a negative example
Minimally specialize the most general hypothesis set
Specialization
set of aa aa na ab 4a nb a 4b n 4

30. 5 is negative(cont’d)
Minimally specialize the most general hypothesis set aa na ab 4a nb a 4b n 4

31. After 3 examples (2 positive,1 negative)
G and S, and all hypotheses in between form exactly the version space ab nb a
1. If an hypothesis between G and S disagreed with an example x, then an hypothesis G or S would also disagree with x, hence would have been removed n

32. After 3 examples (2 positive,1 negative)
G and S, and all hypotheses in between form exactly the version space ab nb a
2. If there were an hypothesis not in this set which agreed with all examples, then it would have to be either no more specific than any member of G – but then it would be in G – or no more general than some member of S – but then it would be in S n

33. Do 8, 6, j satisfy CONCEPT?
At this stage ab nb a No n
Yes
Maybe
Do 8, 6, j satisfy CONCEPT?

34. 2 is the next positive example
Minimally generalize the most specific hypothesis set ab nb a n

35. j is the next negative example
Minimally specialize the most general hypothesis set ab nb

36. Result
+ 4 7 2
– 5 j nb
NUM(r)  BLACK(s)  REWARD([r,s])

37. The version space algorithm
Begin
Initialize S to the first positive training instance N is the set of all negative instances seen so far;
For each example x
If x is positive, then (G,S)  POSITIVE-UPDATE(G,S,x)
else (G,S)  NEGATIVE-UPDATE(G,S,x)
If G = S and both are singletons, then the algorithm has found a single concept that is consistent with all the data and the algorithm halts
If G and S become empty, then there is no concept that covers all the positive instances and none of the negative instances
End

38. The version space algorithm (cont’d)
POSITIVE-UPDATE(G,S,x)
Begin
Delete all members of G that fail to match x
For every s  S, if s does not match x, replace s with its most specific generalizations that match x;
Delete from S any hypothesis that is more general than some other hypothesis in S;
Delete from S any hypothesis that is neither more specific than nor equal to a hypothesis in G; (different than the textbook)
End;

39. The version space algorithm (cont’d)
NEGATIVE-UPDATE(G,S,x)
Begin
Delete all members of S that match x
For every g  G, that matches x, replace g with its most general specializations that do not match x;
Delete from G any hypothesis that is more specific than some other hypothesis in G;
Delete from G any hypothesis that is neither more general nor equal to hypothesis in S; (different than the textbook)
End;

40. Comments on Version Space Learning (VSL)
It is a bi-directional search. One direction is specific to general and is driven by positive instances. The other direction is general to specific and is driven by negative instances.
It is an incremental learning algorithm. The examples do not have to be given all at once (as opposed to learning decision trees.) The version space is meaningful even before it converges.
The order of examples matters for the speed of convergence
As is, cannot tolerate noise (misclassified examples), the version space might collapse

41. Examples and near misses for the concept “arch”

42. More on generalization operators
Replacing constants with variables. For example, color (ball,red) generalizes to color (X,red)
Dropping conditions from a conjunctive expression. For example, shape (X, round)  size (X, small)  color (X, red) generalizes to shape (X, round)  color (X, red)

43. More on generalization operators (cont’d)
Adding a disjunct to an expression. For example, shape (X, round)  size (X, small)  color (X, red) generalizes to shape (X, round)  size (X, small)  ( color (X, red)  (color (X, blue) )
Replacing a property with its parent in a class hierarchy. If we know that primary_color is a superclass of red, then color (X, red) generalizes to color (X, primary_color)

44. Another example sizes = {large, small} colors = {red, white, blue}
shapes = {sphere, brick, cube}
object (size, color, shape)
If the target concept is a “red ball,” then size should not matter, color should be red, and shape should be sphere
If the target concept is “ball,” then size or color should not matter, shape should be sphere.

45. A portion of the concept space

46. Learning the concept of a “red ball”
G : { obj (X, Y, Z)} S : { }
positive: obj (small, red, sphere)
G: { obj (X, Y, Z)} S : { obj (small, red, sphere) }
negative: obj (small, blue, sphere)
G: { obj (large, Y, Z), obj (X, red, Z), obj (X, white, sphere) obj (X,Y, brick), obj (X, Y, cube) } S: { obj (small, red, sphere) } delete from G every hypothesis that is neither more general than nor equal to a hypothesis in S
G: {obj (X, red, Z) } S: { obj (small, red, sphere) }

47. Learning the concept of a “red ball” (cont’d)
G: { obj (X, red, Z) } S: { obj (small, red, sphere) }
positive: obj (large, red, sphere)
G: { obj (X, red, Z)} S : { obj (X, red, sphere) }
negative: obj (large, red, cube)
G: { obj (small, red, Z), obj (X, red, sphere), obj (X, red, brick)} S: { obj (X, red, sphere) } delete from G every hypothesis that is neither more general than nor equal to a hypothesis in S
G: {obj (X, red, sphere) } S: { obj (X, red, sphere) } converged to a single concept

48. LEX: a program that learns heuristics
Learns heuristics for symbolic integration problems
Typical transformations used in performing integration include OP1:  r f(x) dx  r  f(x) dx OP2:  u dv  uv -  v du OP3: 1 * f(x)  f(x) OP4:  (f1(x) + f2(x)) dx   f1(x) dx +  f2(x) dx
A heuristic tells when an operator is particularly useful: If a problem state matches  x transcendental(x) dx then apply OP2 with bindings u = x dv = transcendental (x) dx

49. A portion of LEX’s hierarchy of symbols

50. The overall architecture
A generalizer that uses candidate elimination to find heuristics
A problem solver that produces positive and negative heuristics from a problem trace
A critic that produces positive and negative instances from a problem traces (the credit assignment problem)
A problem generator that produces new candidate problems

51. A version space for OP2 (Mitchell et al.,1983)

52. Comments on LEX
The evolving heuristics are not guaranteed to be admissible. The solution path found by the problem solver may not actually be a shortest path solution.
The problem generator is the least developed part of the program.
Empirical studies: before: 5 problems solved in an average of 200 steps train with 12 problems after: 5 problems solved in an average of 20 steps

53. More comments on VSL Still lots of research going on
Uses breadth-first search which might be inefficient:
might need to use beam-search to prune hypotheses from G and S if they grow excessively
another alternative is to use inductive-bias and restrict the concept language
How to address the noise problem? Maintain several G and S sets.

54. Decision Trees
A decision tree allows a classification of an object by testing its values for certain properties
check out the example at:
The learning problem is similar to concept learning using version spaces in the sense that we are trying to identify a class using the observable properties.
It is different in the sense that we are trying to learn a structure that determines class membership after a sequence of questions. This structure is a decision tree.

55. Reverse engineered decision tree of the whale watcher expert system
see flukes?
yes no
see dorsal fin? no
yes
(see next page)
size?
size med?
vlg
med
yes no
blue
whale
blow
forward?
blows?
Size?
yes no 1 2 lg
vsm
sperm
whale
humpback
whale
gray
whale
right
whale
bowhead
whale
narwhal
whale

56. Reverse engineered decision tree of the whale watcher expert system (cont’d)
see flukes?
yes no
see dorsal fin? no
(see previous page)
yes
blow?
yes no
size? lg sm
dorsal fin and
blow visible
at the same time?
dorsal fin
tall and pointed?
yes no
yes no
killer
whale
northern
bottlenose
whale
sei
whale
fin
whale

57. What does the original data look like?

58. The search problem
Given a table of observable properties, search for a decision tree that
correctly represents the data (assuming that the data is noise-free), and
is as small as possible.
What does the search tree look like?

59. Comparing VSL and learning DTs
A hypothesis learned in VSL can be represented as a decision tree.
Consider the predicate that we used as a VSL example: NUM(r)  BLACK(s)  REWARD([r,s])
The decision tree on the right represents it:
NUM?
True
False
BLACK?
False
False
True
True
False

60. Predicate as a Decision Tree
The predicate CONCEPT(x)  A(x)  (B(x) v C(x)) can
be represented by the following decision tree:
Example: A mushroom is poisonous iff it is yellow and small, or yellow,
big and spotted
x is a mushroom
CONCEPT = POISONOUS
A = YELLOW
B = BIG
C = SPOTTED
D = FUNNEL-CAP
E = BULKY A? B? C?
True
False

61. Training Set True False 13 12 11 10 9 8 7 6 5 4 3 2 1 CONCEPT E D C B
Ex. #

62. Possible Decision TreeA T F

63. Possible Decision TreeA T F
CONCEPT 
(D  (E v A)) v (C  (B v ((E  A) v A))) A? B? C?
True
False
CONCEPT  A  (B v C)
KIS bias  Build smallest decision tree
Computationally intractable problem  greedy algorithm

64. Getting Started The distribution of the training set is:
True: 6, 7, 8, 9, 10,13
False: 1, 2, 3, 4, 5, 11, 12

65. Getting Started The distribution of training set is:
True: 6, 7, 8, 9, 10,13
False: 1, 2, 3, 4, 5, 11, 12
Without testing any observable predicate, we
could report that CONCEPT is False (majority rule) with an estimated probability of error P(E) = 6/13

66. Getting Started The distribution of training set is:
True: 6, 7, 8, 9, 10,13
False: 1, 2, 3, 4, 5, 11, 12
Without testing any observable predicate, we
could report that CONCEPT is False (majority rule) with an estimated probability of error P(E) = 6/13
Assuming that we will only include one observable
predicate in the decision tree, which predicate should we test to minimize the probability of error?

67. Assume It’s A A F T True: 6, 7, 8, 9, 10, 13 False: 11, 12
1, 2, 3, 4, 5 T F
If we test only A, we will report that CONCEPT is True if A is True (majority rule) and False otherwise
The estimated probability of error is:
Pr(E) = (8/13)x(2/8) + (5/8)x0 = 2/13

68. Assume It’s B B F T True: 9, 10 6, 7, 8, 13 False: 2, 3, 11, 12
1, 4, 5 T F
If we test only B, we will report that CONCEPT is False if B is True and True otherwise
The estimated probability of error is:
Pr(E) = (6/13)x(2/6) + (7/13)x(3/7) = 5/13
6, 7, 8, 13

69. Assume It’s C C F T True: 6, 8, 9, 10, 13 7 False: 1, 3, 4
1, 5, 11, 12 T F
If we test only C, we will report that CONCEPT is True if C is True and False otherwise
The estimated probability of error is:
Pr(E) = (8/13)x(3/8) + (5/13)x(1/5) = 4/13 7

70. Assume It’s D D F T True: 7, 10, 13 6, 8, 9 False: 3, 5
If we test only D, we will report that CONCEPT is True if D is True and False otherwise
The estimated probability of error is:
Pr(E) = (5/13)x(2/5) + (8/13)x(3/8) = 5/13
True:
False:
7, 10, 13
3, 5
1, 2, 4, 11, 12
6, 8, 9

71. So, the best predicate to test is A
Assume It’s E E
True:
False:
8, 9, 10, 13
1, 3, 5, 12
2, 4, 11 T F
If we test only E we will report that CONCEPT is False,
independent of the outcome
The estimated probability of error is unchanged:
Pr(E) = (8/13)x(4/8) + (5/13)x(2/5) = 6/13
6, 7
So, the best predicate to test is A

72. Choice of Second Predicate
False C F T
True:
False:
6, 8, 9, 10, 13
11, 12 7
The majority rule gives the probability of error Pr(E|A) = 1/8 and Pr(E) = 1/13

73. Choice of Third Predicate
False F T
True B T F
True:
False:
11,12 7

74. Final Tree L  CONCEPT  A  (C v B) A? B? C? A C B True False True

75. Learning a decision tree
Function induce_tree (example_set, properties)
begin if all entries in example_set are in the same class then return a leaf node labeled with that class else if properties is empty then return leaf node labeled with disjunction of all classes in example_set else begin select a property, P, and make it the root of the current tree; delete P from properties; for each value, V, of P begin create a branch of the tree labeled with V; let partitionv be elements of example_set with values V for property P; call induce_tree (partitionv, properties), attach result to branch V end end end
If property V is Boolean: the partition will contain two
sets, one with property V true and one with false

76. What happens if there is noise in the training set?
The part of the algorithm shown below handles this:
if properties is empty then return leaf node labeled with disjunction of all classes in example_set
Consider a very small (but inconsistent) training set:
A classification T T F F F T A?
True
False
False 
True
True

77. Using Information Theory
Rather than minimizing the probability of error, most existing learning procedures try to minimize the expected number of questions needed to decide if an object x satisfies CONCEPT.
This minimization is based on a measure of the “quantity of information” that is contained in the truth value of an observable predicate and is explained in Section We will skip the technique given there and use the “probability of error” approach.

78. Assessing performance
size of training set
% correct on test set
100
Typical learning curve

79. The evaluation of ID3 in chess endgame

80. Other issues in learning decision trees
If data for some attribute is missing and is hard to obtain, it might be possible to extrapolate or use “unknown.”
If some attributes have continuous values, groupings might be used.
If the data set is too large, one might use bagging to select a sample from the training set. Or, one can use boosting to assign a weight showing importance to each instance. Or, one can divide the sample set into subsets and train on one, and test on others.

81. Explanation based learning
Idea: can learn better when the background theory is known
Use the domain theory to explain the instances taught
Generalize the explanation to come up with a “learned rule”

82. Example
We would like the system to learn what a cup is, i.e., we would like it to learn a rule of the form: premise(X)  cup(X)
Assume that we have a domain theory: liftable(X)  holds_liquid(X)  cup(X) part (Z,W)  concave(W)  points_up  holds_liquid (Z) light(Y)  part(Y,handle)  liftable (Y) small(A)  light(A) made_of(A,feathers)  light(A)
The training example is the following: cup (obj1) small(obj1) small(obj1) part(obj1,handle) owns(bob,obj1) part(obj1,bottom) part(obj1, bowl) points_up(bowl) concave(bowl) color(obj1,red)

83. First, form a specific proof that obj1 is a cup
cup (obj1)
liftable (obj1)
holds_liquid (obj1)
light (obj1)
part (obj1, handle)
part (obj1, bowl)
points_up(bowl)
concave(bowl)
small (obj1)

84. Second, analyze the explanation structure to generalize it

85. Third, adopt the generalized the proof
cup (X)
liftable (X)
holds_liquid (X)
light (X)
part (X, handle)
part (X, W)
points_up(W)
concave(W)
small (X)

86. The EBL algorithm Initialize hypothesis = { }
For each positive training example not covered by hypothesis:
1. Explain how training example satisfies target concept, in terms of domain theory
2. Analyze the explanation to determine the most general conditions under which this explanation (proof) holds
3. Refine the hypothesis by adding a new rule, whose premises are the above conditions, and whose consequent asserts the target concept

87. Wait a minute!
Isn’t this “just a restatement of what the learner already knows?”
Not really
a theory-guided generalization from examples
an example-guided operationalization of theories
Even if you know all the rules of chess you get better if you play more
Even if you know the basic axioms of probability, you get better as you solve more probability problems

88. Comments on EBL
Note that the “irrelevant” properties of obj1 were disregarded (e.g., color is red, it has a bottom)
Also note that “irrelevant” generalizations were sorted out due to its goal-directed nature
Allows justified generalization from a single example
Generality of result depends on domain theory
Still requires multiple examples
Assumes that the domain theory is correct (error-free)---as opposed to approximate domain theories which we will not cover.
This assumption holds in chess and other search problems.
It allows us to assume explanation = proof.

89. Two formulations for learning
Inductive
Given:
Instances
Hypotheses
Target concept
Training examples of the target concept
Analytical
Given:
Instances
Hypotheses
Target concept
Training examples of the target concept
Domain theory for explaining examples
Determine:
Hypotheses consistent with the training examples
Determine:
Hypotheses consistent with the training examples and the domain theory

90. Two formulations for learning (cont’d)
Inductive
Hypothesis fits data
Statistical inference
Requires little prior knowledge
Syntactic inductive bias
Analytical
Hypothesis fits domain theory
Deductive inference
Learns from scarce data
Bias is domain theory
DT and VS learners are “similarity-based”
Prior knowledge is important. It might be one of the reasons for humans’ ability to generalize from as few as a single training instance.
Prior knowledge can guide in a space of an unlimited number of generalizations that can be produced by training examples.

91. An example: META-DENDRAL
Learns rules for DENDRAL
Remember that DENDRAL infers structure of organic molecules from their chemical formula and mass spectrographic data.
Meta-DENDRAL constructs an explanation of the site of a cleavage using
structure of a known compound
mass and relative abundance of the fragments produced by spectrography
a “half-order” theory (e.g., double and triple bonds do not break; only fragments larger than two carbon atoms show up in the data)
These explanations are used as examples for constructing general rules

92. Analogical reasoning
Idea: if two situations are similar in some respects, then they will probably be in others
Define the source of an analogy to be a problem solution. It is a theory that is relatively well understood.
The target of an analogy is a theory that is not completely understood.
Analogy constructs a mapping between corresponding elements of the target and the source.

94. Example: atom/solar system analogy
The source domain contains: yellow(sun) blue(earth) hotter-than(sun,earth) causes(more-massive(sun,earth), attract(sun,earth)) causes(attract(sun,earth), revolves-around(earth,sun))
The target domain that the analogy is intended to explain includes: more-massive(nucleus, electron) revolves-around(electron, nucleus)
The mapping is: sun  nucleus and earth  electron
The extension of the mapping leads to the inference: causes(more-massive(nucleus,electron), attract(nucleus,electron)) causes(attract(nucleus,electron), revolves-around(electron,nucleus))

95. A typical framework
Retrieval: Given a target problem, select a potential source analog.
Elaboration: Derive additional features and relations of the source.
Mapping and inference: Mapping of source attributes into the target domain.
Justification: Show that the mapping is valid.