Mathematical Induction

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Mathematical Induction
Computational Logic Lecture 16
Michael Genesereth Spring 2004
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Incomplete Induction
Definition f(1)1 f(x1)f(x)2x1 Data
Conjecture In this case, the answer is
correct. Lucky Guess.
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Not So Lucky Guess
Data Theorem by Fermat (1601-1665) Fac
t discovered (mercifully) after his death Oops.
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Basic Metavocabulary
Data Base Case Prove for n1 f(1)112 Indu
ctive Case Assume true for x prove for
x1 f(x1)f(x)2x1 f(x1)x22x1 f(x
1)(x1)2 Jules Henri Poincare (1854-1912)
credited with invention.
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Outline
Linear Induction Input successor function on
individuals Output universally quantified
conclusions Structural Induction Input
constructor function for structures Output
universally quantified conclusions Ordered
Induction Input ordering relation on
individuals Output universally quantified
conclusions
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Linear Induction
Linearly Structured World In other words,
there is a distinguished base element and there
is a successor function, which, starting at the
base element, enumerates all elements in the
universe of discourse. Base element
e Successor function f
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Linear Induction Schema
If a property holds of the base element and if it
holds of a successor whenever it holds of an
element, we would like to assert that it holds of
all elements in the universe of discourse. ?e
??x.(?x??f(x)) ? ?x.?x Base case
?e Inductive case ?x.(?x??f(x))
Inductive antecedent ?x Inductive
consequent ?f(x)) Conclusion
?x.?x Analogy to students in class passing on
messages
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Arithmetic Examples
Object constant 0 Unary function constant s
(1) Binary relation constants even, odd,
integer Induction Schema ?0
??x.(?x??s(x)) ? ?x.?x Instance of
Induction Schema integer(0)??x.(integer(x)?
integer(s(x)))??x.integer(x) Instance of
Induction Schema (even(0)?odd(0))
??x.(even(x)?odd(x)?even(s(x))?odd(s(x)))
??x.(even(x)?odd(x))
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Arithmetic Problem
Object constant 0 Unary function constant s
(1) Binary relation constants
integer Axioms integer(0) ?x.(integer(x)?
integer(s(x))) Goal ?x.integer(x) Induction
Schema integer(0)??x.(integer(x)?
integer(s(x)))??x.integer(x)
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Clausal Form
integer(0)??x.(integer(x)?
integer(s(x)))??x.integer(x) I
?(integer(0)??x.(?integer(x)?integer(s(x))))??x.in
teger(x) N ?integer(0)???x.(?integer(x)?integer(s
(x)))??x.integer(x) ?integer(0)??x.?(?integer
(x)?integer(s(x)))??x.integer(x)
?integer(0)??x.(??integer(x)??integer(s(x)))??x.in
teger(x) ?integer(0)??x.(integer(x)??integer(
s(x)))??x.integer(x) S E ?integer(0)?(integer(
a)??integer(s(a)))??x.integer(x) A
?integer(0)?(integer(a)??integer(s(a)))?integer(x)
D ?integer(0)?integer(a)?integer(x)
?integer(0)??integer(s(a))?integer(x) O
?integer(0), integer(a), integer(x)
?integer(0), ?integer(s(a)), integer(x)
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Resolution Proof
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Arithmetic Problem
Object constant 0 Unary function constant
s Binary relation constants even,
odd Axioms even(0) ?x.(even(x)?
odd(s(x))) ?x.(odd(x)? even(s(x))) Goal ?x.(even(
x) ?odd(x)) Induction Schema (even(0)?odd(0))
??x.(even(x)?odd(x)?even(s(x))?odd(s(x)))
??x.(even(x)?odd(x))
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Clausal Form
(even(0)?odd(0)) ??x.(even(x)?odd(x)?even(s(x)
)?odd(s(x))) ??x.(even(x)?odd(x))
?even(0), even(a), odd(a), even(x), odd(x)
?odd(0), even(a), odd(a), even(x), odd(x)
?even(0), ?even(s(a)), even(x), odd(x)
?even(0), ?odd(s(a)), even(x), odd(x)
?odd(0), ?even(s(a)), even(x), odd(x)
?odd(0), ?odd(s(a)), even(x), odd(x)
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Induction and Resolution
Understandability Okay for computers
Terrible for humans Generality Induction
is an axiom schema. Resolution works on
sentences.
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Linear Induction Method
Using the Induction schema to prove a universally
quantified formula. (1) Base Case. Prove the
base case. (2) Inductive Case. Assume ground
version of induction antecedent (induction
hypothesis) and prove corresponding version of
induction consequent. If successful, the
universally quantified conclusion holds. Why?
Deduction Theorem.
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Even and Odd
Axioms even(0) even(x) ? odd(s(x)) odd(x)
? even(s(x)) Desired Conclusion ?x.(even(x)?odd
(x)) Induction Axiom (even(0)?odd(0))
??x.(even(x)?odd(x)?even(s(x))?odd(s(x)))
??x.(even(x)?odd(x))
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Even and Odd Problem
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Binary Addition
Object constant 0 Unary function constant
s Binary function constant Chain
Axioms s(x)s(y) ? xy Addition
Axioms x0x xs(y)s(xy) s(x)ys(xy)
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Binary Addition Problem
Question ?x.0xx Axioms s(x)s(y) ?
xy x0x xs(y)s(xy) s(x)ys(xy) Goal ?x.0
xx Induction Axiom 000 ??x.(0xx ?
0s(x)s(x)) ? ?x.0xx
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Binary Addition Problem
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Binary Addition Problem
Axioms s(x)s(y) ? xy x0x xs(y)s(xy) s(x)y
s(xy) Question ?x.xyyx Induction
Axiom 0yy0 ??x.(xyyx ? s(x)yys(x)) ?
?x.xyyx
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Binary Addition Problem
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Intuition for Structural Induction
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Binary Tree Representation
Tree Representation as a
term pair(pair(a,b),pair(pair(c,d),pair(e,f))) I
nfix version (ab)((cd)(ef))
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Induction and Resolution
Base elements a, b Successor function
Induction Schema ?a ??b ??x.?y.(?x
??y ??xy) ? ?x.?x
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Reverse
Object Constants a, b Unary Function Constant
rev Binary Function Constant Axioms rev(a)
a rev(b)b rev(xy)rev(y)rev(x)
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Reverse Problem
Desired Conclusion rev(rev(x))x Axioms rev(
a)a rev(b)b rev(xy)rev(y)rev(x) Induct
ion Axiom rev(rev(a))a?rev(rev(b))b
??x.?y.(rev(rev(x))x?rev(rev(y))y ?
rev(rev(xy))xy) ? ?x. rev(rev(x))x
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Reverse Problem
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Ordered Induction
Minimal Element e Ordering Relation
lt Inductive Formula ? Induction
Schema ?x.(?y.(yltx??y)??x) ??x.?x
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Base Case
Ordered Induction Schema ?x.(?y.(yltx??y)??x)
??x.?x It appears that there is no base case.
Ah, but there is! Remember that e is a minimal
element. ?x.?xlte Let x be e in induction
antecedent. ?y.(ylte??y)??e) Antecedent of
this sentence must always be true. Therefore, to
prove the condition, necessary to prove ?e.
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Goldbachs Conjecture
Data Conjecture (1742) As of late
60's, still not proved. Single counterexample
would disprove.
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Summary
The key to induction is having some local way of
working through space of all objects and using
that to establish arbitrary formulas. Induction
Method Instance of schema can be used by
standard proof methods. Induction schema cannot
be written as finite set of axioms. Can state
as a metalevel axiom and use metareasoning.
Induction method works when schema true, and it's
simple.