CM10196
Topic 3: Proof by Induction
Guy McCusker
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The Mathematics of Infinity
One of the most powerful and fascinating aspects of mathematics is the
ability to reason about infinite objects.
Examples of simple infinite objects that we can “tame” using simple
mathematical techniques include
I
the set N of natural numbers 0, 1, 2, . . .
I
the set of all boolean formulae
I
the set of all elements of a given data structure: all lists, all trees, etc.
The technique we will use to reason about these infinite sets is
mathematical induction.
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What’s the problem?
Suppose we want to prove some property about every natural number. For
instance
every natural number greater than 1 is either a prime, or divisible
by a prime.
That is,
∀n ∈ N.n > 1 → (“n is prime” ∨ “n is divisible by a prime”)
Since this statement says something about every natural number, proving
that it is true could be hard work. A simple-minded approach would try to
prove that it’s true for 0, and for 1, and for 2, and for 3, and for 4. . .
That will take forever, which is too long.
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Induction
Mathematical induction lets us prove a statement about all natural
numbers by proving only two facts:
I
the statement is true of the number 0
I
if the statement is true of some number n, then it is true of n + 1.
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Like toppling dominoes
Suppose we can establish that these two statements are true:
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The first domino (which we call domino 0) is going to fall.
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If domino n is going to fall, so is domino n + 1.
Then we can conclude that all the dominoes are going to fall.
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Proving that the dominoes will fall
How can we prove that all the dominoes will fall?
I
Establish directly that the first one is going to fall. For example, we
could just knock it over ourselves!
I
Establish that the dominoes are sufficiently close together that when
one falls, the next one does.
The second point is essentially saying that the dominoes have a particular
property. It is this property that makes them all fall down.
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Never mind the dominoes, here’s the natural numbers
Here’s a property of the natural numbers:
∀n ∈ N.n3 − n is divisible by 6.
To prove this by induction, we need to prove:
I
0 satisfies the property, i.e. 03 − 0 is divisible by 6.
I
if some n satisfies the property then so does n + 1, i.e. if n3 − n is
divisible by 6 then so is (n + 1)3 − (n + 1).
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Why will this work?
If we can prove these two things, then we will know that every n3 − n is
divisible by 6. Why?
I
We know that 03 − 0 is divisible by 6: that was the first thing we’ve
proved.
I
We know that if 03 − 0 is divisible by 6, so is 13 − 1. Putting these
two things together we can conclude that 13 − 1 is divisible by 6.
I
We know that if 13 − 1 is divisible by 6, so is 23 − 2. Putting this
together with the last fact, we can conclude that 23 − 2 is divisible by
6.
I
We know that if 23 − 2 is divisible by 6, so is 33 − 3. Putting this
together with the last fact, we can conclude that 33 − 3 is divisible by
6.
I
...
That’s why proof by induction is a valid principle.
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Let’s do the proof
For the first part, called the “base case” or the “basis” of the induction,
we simply notice that 03 − 0 = 0 which is divisible by 6.
The second part requires more work.
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The inductive step
For the second part, called the “inductive step”, we reason as follows.
(n + 1)3 − (n + 1) = n3 + 3n2 + 3n + 1 − n − 1 = (n3 − n) + 3n2 + 3n
If n3 − n is divisible by 6, then this number will be divisible by 6 as long as
3n2 + 3n
is divisible by 6. But
3n2 + 3n = 3n(n + 1)
which will be divisible by 6 as long as one of n or n + 1 is divisible by 2.
For any natural number n, either n is divisible by 2 or n + 1 is, so our
proof is complete.
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Writing down an inductive proof
Inductive proofs are normally laid out in two paragraphs, one for the basis
and one for the inductive step. Like this.
Basis. We must show that 03 − 0 = 0 is divisible by 6, which is trivially
true.
Inductive step. We must show that (n + 1)3 − (n + 1) is divisible by 6 if
n3 − n is. But
(n + 1)3 − (n + 1) = (n3 − n) + 3n(n + 1).
We know n3 − n is divisible by 6 by the inductive hypothesis, so it suffices
to show that 3n(n + 1) is divisible by 6, i.e. that n(n + 1) is divisible by 2.
But one of n and n + 1 must be even, so this is the case.
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Inductive hypothesis
In the inductive step, we only have to show that (n + 1)3 − (n + 1) is
divisible by 6 if n3 − n is.
That means we can use the fact that n3 − n is divisible by 6 in our
argument that (n + 1)3 − (n + 1) is.
When we do this, we are “using the inductive hypothesis”.
This is what makes inductive proofs work.
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A classic example: summing the naturals
For every natural number n,
x=n
X
x=
x=0
n(n + 1)
.
2
We prove this by induction.
Basis. We must prove that the sum of naturals from 0 to 0 is equal to
0(0 + 1)/2 = 0, which is true.
Inductive step. We must prove that the sum of naturals from 0 to n + 1 is
n(n + 1) 2(n + 1)
n(n + 1)
(n + 1)(n + 2)
=
+
=
+ (n + 1).
2
2
2
2
Px=n
By the inductive hypothesis, n(n + 1)/2 = x=0 x so the expression
above is equal to
!
x=n
X
x + (n + 1)
x=0
which is clearly the required sum.
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The principle of induction as a logical formula
We can formulate the principle of induction as a logical formula: if P(n) is
any predicate on the natural numbers, the principle of induction states that
P(0) ∧ ∀n ∈ N.(P(n) → P(n + 1)) → ∀n ∈ N.P(n).
I
Basis
I
Inductive step
I
Conclusion
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Proving the inductive step
The inductive step requires us to prove that
∀n ∈ N.(P(n) → P(n + 1))
is true.
The way we do this in practice is:
I
let n be any natural number
I
show that P(n + 1) holds provided P(n) does.
What this means is we show that P(n + 1) holds using the “fact” that
P(n) holds.
We don’t know that P(n) is true; we just show that, if it is, then so is
P(n + 1).
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Example: the size of powersets
If A is a finite set, we write |A| for the number of elements in A.
Let’s prove that
|PA| = 2|A| .
What we will show is in fact this:
For all natural numbers n, for all B ⊆ A such that |B| = n, |PB| = 2n .
See how we’ve turned a property of sets into something quantified over all
natural numbers. . .
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Size of powersets
Basis. We must show that for every B ⊆ A with |B| = 0, it is the case
that |PB| = 20 = 1.
The only set with size 0 is ∅, the empty set. Its powerset is {∅}, which has
size 1, so the base case is complete.
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Size of powersets
Inductive step. We must show that if the property holds for all B ⊆ A
with |B| = n, then it also holds for those B ⊆ A with |B| = n + 1.
Suppose |B| = n + 1. Since this is at least 1, there must be an element
b ∈ B.
Consider the set B \ {b}. This is a subset of A and its size is n. Therefore
by the inductive hypothesis
|P(B \ {b})| = 2n .
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Size of powersets
What subsets of B are there? To help us analyze them, let’s consider a
small example.
If B is the set {a, b, c}, then the subsets of B are:
I those that do not contain b:
1
2
3
4
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∅
{a}
{c}
{a, c}
those that do contain b:
1
2
3
4
{b} = {b} ∪ ∅
{a, b} = {b} ∪ {a}
{b, c} = {b} ∪ {c}
{a, b, c} = {b} ∪ {a, c}
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Inductive step
In general, there are
I
subsets that do not contain b
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subsets that contain b.
The subsets that do not contain b are exactly the same as the subsets of
B \ {b}, so there are 2n of them.
Every subset that does contain b is of the form {b} ∪ S, where S is a
subset that does not contain b. So, there are 2n of these too.
So in total there are 2n + 2n = 2 × 2n = 2n+1 subsets of B.
The inductive step is complete.
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Concluding
We’ve shown that for every subset B of A, if |B| = n then |PB| = 2n .
A is a subset of itself, so we can conclude that
|PA| = 2|A|
as required.
What we’ve done here is to turn a statement about sets into a statement
about natural numbers by measuring something in our original statement.
We then use induction on this measurement to get our proof done.
In this case, we say we’ve proved the statement by induction on the size of
A.
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Different base cases
If we’re only interested in showing that P(n) holds for n > 5, for instance,
we can set our induction up slightly differently:
Basis. Show that P(6) holds.
Inductive step. Show that if P(n) holds for some n > 5 then P(n + 1)
holds.
Exercise
Give a formula that expresses this principle, and show that it follows from
the standard principle of induction. Hint: given a predicate P(n), consider
the predicate P 0 (n) defined to be P(n + 6), i.e. P 0 (n) holds if P(n + 6)
holds.
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A logical example: generalized distributive law
Show that, for any formulae x and y1 , . . . , yn , (with n ≥ 1),
x ∧ (y1 ∨ · · · ∨ yn ) ≡ (x ∧ y1 ) ∨ · · · ∨ (x ∧ yn ).
Basis. The base case here is n = 1, and in that case both sides of the
equivalence are x ∧ y1 so the equivalence holds.
Inductive step.
x ∧ (y1 ∨ · · · ∨ yn ∨ yn+1 ) ≡ x ∧ ((y1 ∨ · · · ∨ yn ) ∨ yn+1 )
using associativity of ∨
≡ (x ∧ (y1 ∨ · · · ∨ yn )) ∨ (x ∧ yn+1 )
using distributivity
≡ (x ∧ y1 ) ∨ · · · ∨ (x ∧ yn ) ∨ (x ∧ yn+1 )
using the inductive hypothesis.
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Divisibility by primes
We motivated the need for the principle of induction with this example:
we’d like to prove that
every natural number greater than 1 is either a prime, or divisible
by a prime.
Let’s try to prove it by induction.
Basis. This is the case n = 2. It is prime, so it satisfies the property.
Inductive step. We must prove that the property holds for n + 1 if it does
for n. The number n + 1 is either prime or not. If it is, then it satisfies the
property. If it’s not, then. . . what?
We know it must be divisible by something. If it’s divisible by n then we
can use our inductive hypothesis: n is itself a prime, or divisible by a
prime. So if n + 1 is divisible by n, then n + 1 is divisible by a
prime—either n itself or the prime that n is divisible by.
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Uh oh!
But wait!
We don’t know that n + 1 is divisible by n, and in fact it won’t be, except
when n is 1. (Can you see why?)
So we can’t use our inductive hypothesis. Oh no!
What we know is that either n + 1 is prime (which would mean we have
proved what we need) or it’s divisible by some number m. We also know
that m ≤ n.
If we could apply our inductive hypothesis to m, we’d be fine: the
reasoning from the last slide would work. But the principle of induction
only gives us an inductive hypothesis for the number n, not for every
m ≤ n.
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Course-of-values induction aka strong induction
It seems like there should be a way to rescue the situation. After all, by
the time domino n + 1 is ready to fall, not only has domino n fallen, all
dominoes up to n, including domino m, have fallen.
That is to say, our property should be true of every m ≤ n.
To put it another way, we should be able to use our inductive hypothesis
on every m ≤ n.
The principle of course-of-values induction, also known as strong
induction, allows just this.
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Course-of-values induction
Course-of-values induction tells us that, to prove that a property P(n)
holds for every n ∈ N, it’s enough to:
Basis. Prove that P(0) holds, and
Inductive step. Prove that, for any natural number n, if P(m) holds for
every m ≤ n, then P(n + 1) holds.
This gives us a stronger inductive hypothesis to use in the inductive step:
instead of just using the hypothesis that P(n) holds, we can use P(m) for
any m ≤ n.
As with ordinary induction we can use different base values if we only need
a property to hold for numbers greater than a certain minimum.
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Prime divisibility again
We can now prove our statement about the natural numbers using
course-of-values induction.
Basis. We need to show that 2 is either prime or divisible by a prime. It’s
prime so we’re done.
Inductive step. We need to show that, if every m ≤ n has this property,
then so does n + 1.
I
I
If n + 1 is prime, we are done.
If not, n + 1 must be divisible by some m ≤ n. By the inductive
hypothesis, m is either prime or divisible by some prime p.
I
I
If m is prime, we’ve shown that n + 1 is divisible by the prime m, so
we’re done.
If not, m is divisible by p and n + 1 is divisible by m, so n + 1 is
divisible by p, so we’re done.
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Course-of-values induction follows from ordinary induction
The principle of induction is this:
P(0) ∧ ∀n ∈ N.P(n) → P(n + 1) → ∀n ∈ N.P(n).
Course-of-values induction is this:
P(0) ∧ ∀n ∈ N.(∀m ∈ N.m ≤ n → P(m)) → P(n + 1) → ∀n ∈ N.P(n).
It turns out that the course-of-values induction formula follows from the
ordinary induction formula.
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Course-of-values induction from ordinary induction
Take the ordinary induction formula, and replace the predicate P(n) by the
predicate ∀m ∈ N.m ≤ n → P(m).
The base case,
P(0)
becomes
∀m ∈ N.m ≤ 0 → P(m)
which is equivalent to P(0).
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Course-of-values induction from ordinary induction
The inductive step,
∀n ∈ N.P(n) → P(n + 1)
becomes
∀n ∈ N. (∀m ∈ N.m ≤ n → P(m)) → ( ∀m ∈ N.m ≤ n + 1 → P(m) )
This is equivalent to




∀n ∈ N. (∀m ∈ N.m ≤ n → P(m)) → 
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
∀m ∈ N.m ≤ n → P(m)

∧

P(n + 1)
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Simplifying the inductive step
The formula for the inductive step is




∀n ∈ N.  (∀m ∈ N.m ≤ n → P(m)) → 

∀m ∈ N.m ≤ n → P(m)


∧
P(n + 1)
Using the logical law
X → ( X ∧Y) ≡ X → Y
this is equivalent to
∀n ∈ N.
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(∀m ∈ N.m ≤ n → P(m)) → P(n + 1)
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The conclusion
The conclusion
∀n ∈ N.P(n)
becomes
∀n ∈ N.∀m ∈ N.m ≤ n → P(m).
These two statements are logically equivalent.
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Course-of-values induction
Putting all this together, we find that taking the formula for induction:
P(0) ∧ ∀n ∈ N.P(n) → P(n + 1) → ∀n ∈ N.P(n).
and replacing P(n) with ∀m ∈ N.m ≤ n → P(m), we get a formula
equivalent to
P(0) ∧ ∀n ∈ N.(∀m ∈ N.m ≤ n → P(n)) → P(n + 1) → ∀n ∈ N.P(n)
which is our course-of-values induction principle. Phew!
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A stronger fact about primes
Exercise
Use course-of-values induction to show that every natural number can be
written as a product of prime numbers.
That is to say, every natural number n is equal to
p1 × p2 × · · · × pk
for some collection of prime numbers p1 , . . . , pk .
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Inductive definitions
In programming, you have surely seen definitions of functions which look a
little bit like this:
factorial 0 = 1
factorial (n+1) = (factorial n) * (n+1)
How do we know that such a “definition” really does define a function?
Remark
If you reach a stage in your life when you see something like the above and
immediately react by thinking “Is that well-defined?”, you have become a
mathematician!
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Recursion and induction
In programming, we know that we can compute functions like that one
using recursion: when we ask the computer to evaluate
factorial 10
I
it goes off and tries to compute (factorial 9) * 10
I
which makes it compute (factorial 8) * 9 * 10
I
which makes it compute (factorial 7) * 8 * 9 * 10
I
...
I
which makes it compute (factorial 0) * 1 * ...
I
which makes it compute 1 * 1 * ...
* 9 * 10
* 9 * 10
which has a value.
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Recursion and induction
But a recursive program might not terminate, and it will only compute one
possible answer.
In principle, a recursive “function definition” could define nothing at all, or
define a “function” which has more than one possible answer for some
inputs. (This would not actually be a function.)
A mathematician, or a conscientious programmer, would want to be sure
that the definition really did define a function.
Induction helps us do that.
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Well-definedness of factorial
We can prove by induction that for every natural number n, there is a
unique natural number fn such that
f0 = 1
fn+1 = fn × (n + 1)
Basis. We must show that there is a unique number f0 such that f0 = 1.
This is obvious!
Inductive step. We must show that, if there is a unique fn satisfying the
equations, then there is a unique fn+1 . But the second equation tells us
that fn+1 must be fn × (n + 1), so if fn is well-defined, so is fn+1 .
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The Fibonacci sequence
What about a definition like this one?
fib(0) = 1
fib(1) = 1
fib(n+2) = fib(n+1) + fib(n)
How can we be sure that this is well-defined? If we try to use induction as
before, the inductive step looks like this:
Inductive step. We must show that if fib(n) is well defined, so is
fib(n+1). If n = 0 this is easy: fib(1) = 1 by definition. For n > 0, the
definition tells us that fib(n+1) = fib(n) + fib(n-1). Our inductive
hypothesis tells us that fib(n) is well-defined. . .
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Well-definedness of Fibonacci
Uh-oh. Here we need to know that fib(n-1) is well-defined, but the
inductive hypothesis does not help.
The solution should be clear from previous examples: use course-of-values
induction.
In fact we can get away with a slightly weaker form: we only need our
inductive hypothesis to apply to the last two values.
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“Double” induction
Here’s the induction principle we can use to prove that the Fibonacci
sequence is well-defined.
P(0) ∧ P(1) ∧ ∀n ∈ N.(P(n) ∧ P(n + 1)) → P(n + 2) → ∀n ∈ N.P(n).
Exercise
Use this principle to show that the Fibonacci sequence is well-defined.
Harder exercise
Show that this principle follows from the standard induction principle.
Hint: think about how the “base case” of this principle, P(0) ∧ P(1), can
be seen as the base case P 0 (0) of the ordinary induction principle with a
different predicate.
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Another proof of the DNF theorem
In earlier lectures, we gave a slightly informal proof of the fact that every
Boolean formula is equivalent to one in disjunctive normal form.
Now that we have induction to help us, we can give a much more precise
proof.
We will prove the theorem by induction on the size of the formula. That
is, what we are really proving is that
Theorem
For every natural number n, if F is a Boolean formula with n symbols,
then there is a formula F 0 in disjunctive normal form such that F ≡ F 0 .
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DNF and CNF theorem
In fact it turns out to be easier to prove both the DNF and CNF theorems
at the same time, so we’re really going to show that:
Theorem
For every natural number n, if F is a Boolean formula with n symbols,
then there is a formula F 0 in disjunctive normal form such that F ≡ F 0 ,
and a formula F 00 in conjunctive normal form such that F ≡ F 00 .
Our base case will be n = 1, because there are no formulae with zero
symbols. We will use course-of-values induction.
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DNF/CNF theorem
Basis. The only formulae with 1 symbol are variables X , Y etc. These are
already in DNF and in CNF, so we can take F 0 = F 00 = F and we are done.
Inductive step. A Boolean formula with more than one symbol must be
of one of the following forms:
I
(F1 ∧ F2 )
I
(F1 ∨ F2 )
I
(F1 → F2 )
I
¬F .
We will consider these cases one by one.
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DNF/CNF theorem: conjunction case
Consider the case of a formula F = (F1 ∧ F2 ).
The formulae F1 and F2 have fewer symbols than F , so by the inductive
hypothesis, they have DNFs F10 and F20 and CNFs F100 and F200 .
Then the formula (F100 ∧ F200 ) is a CNF which is equivalent to F , so we can
take F 00 to be this, completing the CNF part of the inductive step in this
case.
For the DNF part, we will transform the formula (F10 ∧ F20 ) into a DNF
using the distributive law.
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Distributing
F10 is a DNF so it has the form
G1 ∨ · · · ∨ Gk
where each Gi is a conjunction of literals
Gi = L1 ∧ · · · ∧ Lki .
Similarly, F20 has the form
H1 ∨ · · · ∨ Hl .
Therefore F10 ∧ F20 is equivalent to
(F10 ∧ H1 ) ∨ · · · ∨ (F10 ∧ Hl )
by the generalized distributive law.
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Distributing
Applying distributivity again we find that F10 ∧ F20 is equivalent to
(G1 ∧ H1 ) ∨ (G2 ∧ H1 ) ∨ · · · ∨ (Gk ∧ H1 )
∨
(G1 ∧ H2 ) ∨ (G2 ∧ H2 ) ∨ · · · ∨ (Gk ∧ H2 )
∨
..
.
∨
(G1 ∧ Hl ) ∨ (G2 ∧ Hl ) ∨ · · · ∨ (Gk ∧ Hl )
This is in DNF so this part of the proof is complete.
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DNF/CNF theorem: negation case
Consider the case of a formula F = ¬F1 .
The formula F1 has fewer symbols, so by the inductive hypothesis, it can
be written in both DNF, F10 , and CNF, F100 .
We will use de Morgan’s laws to transform ¬F10 into a CNF, and ¬F200 into
a DNF. In both cases we will end up with a formula equivalent to ¬F1
which is F , so this will complete the proof for this case.
Notice that it’s the DNF of F1 which gives rise to the CNF of F and vice
versa. That is why it is “easier” to prove the DNF and CNF theorems
together.
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DNF/CNF theorem: negation case
If F10 = G1 ∨ · · · ∨ Gk where each Gi has the form L1 ∧ · · · ∧ Lki , then ¬F10
is equivalent to
(¬G1 ∧ · · · ∧ ¬Gk )
using de Morgan’s law.
Using de Morgan’s law again, each ¬Gi is equivalent to something of the
form
¬L1 ∨ · · · ¬Lki .
Finally, where Li = ¬Xi we can replace ¬¬Xi in the above by Xi and
finally we obtain a CNF equivalent to ¬F10 .
Similarly we can use de Morgan’s laws to transform ¬F200 into a DNF.
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DNF/CNF theorem: other cases
The case of a formula of the form (F1 ∨ F2 ) is symmetric to the
conjunction case.
The case of a formula of the form (F1 → F2 ) can be handled by means of
the cases for negation and disjunction, using the fact that
(F1 → F2 ) ≡ (¬F1 ∨ F2 ).
This completes the proof.
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Structural induction
What we did in that proof was very typical of many proofs about things
like formulae or simple data structures in programs:
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we noticed that every formula is built up out of smaller formulae
using certain constructions
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we proved our statement by breaking a formula up into its smaller
formulae and applying the inductive hypothesis to the smaller
formulae.
It seems a bit silly to bother with the business of turning this into an
induction on numbers: talking about the number of symbols in the
formula etc.
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Structural induction
The property of natural numbers that makes induction valid is that if you
take any natural number n and keep subtracting 1 from it, you reach zero
after finitely many steps.
Similarly, for any Boolean formula, if you keep breaking it down into its
subformulae, eventually you reach a collection of variables.
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Structural induction
This gives us an induction principle for Boolean formulae: let P be some
predicate on formulae (e.g. “this formula has a disjunctive normal form”).
We can show that P holds for every formula by showing the following.
Basis. Show that P holds for Boolean variables X .
Inductive step. Show that, if P holds for formulae F1 and F2 then it
holds for (F1 ∧ F2 ), (F1 ∨ F2 ), (F1 → F2 ) and ¬F1 .
(In practice, such proofs look just the same as the one we just wrote for
the DNF/CNF theorem.)
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Why does it work?
This works because, in some sense, Boolean formulae are inductively
defined: something is a Boolean formula if it is
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a variable, X , or
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a formula (F1 ∧ F2 ), (F1 ∨ F2 ), (F1 → F2 ) or ¬F1 where F1 and F2
are Boolean formulae.
This is an inductive definition because larger formulae are defined in terms
of smaller ones.
Whenever you see a collection of things defined in this way, there will be a
structural induction principle that can be used.
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Mathematical induction as structural induction
The ordinary induction principle can also be seen as having this form.
To do this, we have to see the natural numbers as “having structure”
which we can break down.
One important way of defining the natural numbers is as follows. A
natural number is
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The number 0, or
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A number S(n), where n is a natural number.
Here S stands for successor. We think of S(n) as meaning n + 1.
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Structural induction on naturals
The structural induction principle for natural numbers defined like that is
as follows.
To prove that some property P(n) holds for every natural, we need to
prove that
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P(0) holds
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If P(n) holds for some n then P(S(n)) also holds.
In a formula:
P(0) ∧ ∀n ∈ N.P(n) → P(S(n)) → ∀n ∈ N.P(n)
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Peano arithmetic
The idea of defining the natural numbers using 0 and S is used in the
most popular axiomatization of arithmetic, called Peano arithmetic after
its inventor.
Peano arithmetic includes axioms saying things like
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S(n) 6= 0
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S(n) = S(m) → n = m
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P(0) ∧ ∀n ∈ N.P(n) → P(S(n)) → ∀n ∈ N.P(n)
We can then go on to define arithmetic operations like + and ×
inductively. For instance:
a+0 = a
a + S(b) = S(a + b)
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Peano arithmetic
Using only this structural definition, we can establish a lot of facts about
arithmetic quite easily, via induction.
For example, we can show that our inductive (or is it recursive?) definition
of + is commutative. That is, show that for all natural numbers a and b,
a + b = b + a.
Let’s do it!
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Commutativity of addition
Lemma
For any natural a, 0 + a = a.
We prove this by induction on a.
Basis. We must show that 0 + 0 = 0, but that follows from the definition
of +.
Inductive step. We must show that if 0 + a = a then 0 + S(a) = S(a).
But by definition of +, 0 + S(a) = S(0 + a), and by the inductive
hypothesis, 0 + a = a, so our result follows.
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Commutativity of addition
Lemma
For any natural numbers a and b, S(a) + b = S(a + b).
We prove this by induction on b.
Basis. We must show that S(a) + 0 = S(a + 0). By the previous lemma,
S(a) + 0 = S(a), and by definition of +, a + 0 = a so S(a + 0) = S(a) as
required.
Inductive step. We must show that S(a) + S(b) = S(a + S(b)). By
definition of +, S(a) + S(b) = S(S(a) + b) and by the inductive
hypothesis, S(a) + b = S(a + b), so
S(a) + S(b) = S(S(a + b)) = S(a + S(b))
as required.
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Commutativity of addition
Finally we show that a + b = b + a by induction on b.
Basis. We must show that a + 0 = 0 + a. By definition, a + 0 = a, and by
our first lemma, 0 + a = a so we’re done.
Inductive step. We must show that if a + b = b + a then
a + S(b) = S(b) + a. By definition of +, a + S(b) = S(a + b). By our
second lemma, S(b) + a = S(b + a). By our inductive hypothesis,
b + a = a + b so S(b + a) = S(a + b) and the proof is complete.
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More fun with Peano arithmetic
Exercise
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Show that this definition of + satisfies the commutative law.
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Give an inductive definition of the operation of multiplication on
natural numbers. You may use the + operation defined above.
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Show that S(0) is an identity for multiplication, i.e. that
a × S(0) = a.
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Show that multiplication is commutative.
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