3
Mathematical Induction
Reading: Metalogic Part II, 23,27
Contents
3.1 Mathematical Induction . . . . . . . . . . . . . . . . . . . . . . .
3.1.1 Peano’s Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1.2 **Example: Parentheses . . . . . . . . . . . . . . . . . . . . . .
3.1.3 The Principle of Induction on the Construction
of a Wff . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1.4 **The Induction Principle . . . . . . . . . . . . . . . . . . . .
3.2 Unique Decomposition . . . . . . . . . . . . . . . . . . . . . . . .
3.2.1 Unique Readability Theorem . . . . . . . . . . . . . . . . .
3.2.2 Finding the Main Connnective . . . . . . . . . . . . . . . .
3.3 Some Theorems Involving Induction . . . . . . . . . . .
3.3.1 Unique Valuation . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.2 The Interpolation Theorem . . . . . . . . . . . . . . . . . . .
3.3.3 Disjunctive Normal Form . . . . . . . . . . . . . . . . . . . .
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3.1 Mathematical Induction
Our objective in this section is to familiarize the reader with the technique
of Proof by Mathematical Induction. This proof technique is used throughout
mathematical logic, and so we should learn it right away.
3.1.1 Peano’s Axioms
We think of the natural numbers as forming a sequence beginning with 0
followed by its successor, and then its successor, and so on. We begin with 0
and then generate the sequence by adding on n0 whenever we have generated
n. n0 is said to be the successor of n, and it may be thought of as n + 1. We
thus get the sequence
0, 00 , 000 , 0000 , 00000 , . . .
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3 Mathematical Induction
This informal characterization of the set of natural numbers N , is formally
characterized by
Peano’s Postulates1
P1
P2
P3
P4
P5
0 ∈ N.
If n ∈ N , then there is a unique n0 ∈ N .
For all m, n ∈ N , if m0 = n0 , then m = n.
For each n ∈ N , n0 6= 0.
If M ⊆ N such that 0 ∈ M and m0 ∈ M whenever m ∈ M , then
M = N.
P1 is straightforward. P2 tells us that 0 is a function, so that if x = y, then
x0 = y 0 . P3 tells us that the function 0 is 1-1, i.e., that no number can be the
successor of two distinct numbers. P4 tells us that 0 is not the successor of
any number, so that, in effect, the Range of the function 0 is the set N − {0}.
And P5 is the basis for what is known as Weak Induction
Principle 3.1.1 (The Principle of Mathematical Induction) Let P be
a property of natural numbers. Suppose that:
(1) 0 has the property P.
(2) If any natural number n has the property P, then its successor n0 has the
property P.
Then, every natural number n has the property P.
The Principle of Mathematical Induction can be drawn from P5 by letting
M = {m ∈ N |P (m)}.
The following terminology is fairly standard in conjunction with proof by
mathematical induction. The part of the proof which consists in establishing
(1), i.e., that 0 has P, is called the Basis step. The part of the proof which
consists in establishing (2), i.e., that n0 has the property P whenever n has
the property P, is called the Induction Step. The induction step is carried out
by assuming that n has P, and showing that on this assumption n0 has P. This
assumption that n has P is called the Induction Hypothesis.
We will use mathematical induction to prove the following elementary
consequence of these postulates, namely, that every natural number other
than 0 is the successor of a natural number2 :
Proposition 20 Every n ∈ N is such that either n = 0 or else there exists
an m ∈ N and n = m0
Proof By Mathematical Induction.
Basis Step. n = 0. The disjunction follows immediately by addition: either
n = 0 or else there exists an m ∈ N and n = m0 .
Induction Step. Assume the theorem holds for n ∈ N , i.e., assume that
either n = 0 or else there exists an m ∈ N and n = m0 . (This is the Induction
1
2
These are actually due to Dedekind
This establishes that the function 0 is not merely 1-1, but also onto N − {0}
3.1 Mathematical Induction
37
Hypothesis.) We have to show that this also holds for n0 . Clearly n0 ∈ N ,
by P2 . And, clearly n0 6= 0, by P4 . So, we have to show that there exists an
m ∈ N such that n0 = m0 . But, of course, there is: let m = n.
¤
There are a number of variants of this principle that mathematicians rely
on in proofs by induction. Kleene (1950) also acknowledges Course-of-Values
Induction, on which the induction hypothesis is not simply that n has P, but
that m has P, for all m ≤ n. So, we have
Principle 3.1.2 (The Principle of Course-of-Values Induction:) Let P
be a property of natural numbers. Suppose that:
(1) 0 has the property P.
(2) If all natural numbers m ≤ n have the property P, then n0 has the property
P.
Then, every natural number n has the property P.
[Andrews, Mendelson (1987) identifies a Principle of Complete Induction,3
This is also known as Strong Induction:
Principle 3.1.3 (The Principle of Complete Induction (Strong Induction):)
Let P be a property of natural numbers. Suppose that:
If all natural numbers m < n have the property P, then n has the property P.
Then, every natural number n has the property P.
These variations need not upset us, however, for they are provably equivalent. As such, we can choose whichever form makes the proof easiest. We
will show that the Principle of Mathematical Induction is equivalent to the
Principle of Complete Induction in two steps.
Lemma 3.1.1 The Principle of Mathematical Induction is a consequence of
The Principle of Complete Induction.
Proof [Andrews] We want to show that if P (0) and if, for all n, if P (n) then
P (n0 ), then for all n, P (n). So, (as with any conditional proof), we assume
the antecedent and try to derive the consequent. So, we assume
(*) P (0) and for all n, if P (n) then P (n0 )
We want to show that P (n) for all n. We do so using the Principle of Complete
Induction, i.e., we take as our induction hypothesis that P (j) for all j < n,
and try to derive P (n). The Principle of Complete Induction will then entitle
3
Hunter (1996) has a different terminology. “In the Induction Step we prove that
if the Theorem holds for all cases up to an arbitrarily given point, then it holds
also for all cases at the next higher point. (This will be a strong induction. A weak
induction is one in which the Induction Step shows that if the Theorem holds for
all cases at an arbitrarily given point, then it holds also for all cases at the next
higher point.” By these lights, our Principle of Mathematical Induction would be
a weak induction and both our Course-of-Values Induction and our Principle of
Complete Induction would be strong inductions—assuming, of course, that Hunter
would count the Induction Hypothesis in Complete Induction as the Induction
Step even though there is no Basis Step.
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3 Mathematical Induction
us to conclude that P (n) for all n.
Induction Step. Assume P (j) for all j < n. If n = 0, then by (*), P (n).
Otherwise n > 0. So, n = m0 . In that case, m < n, so, P (m) by the induction
hypothesis. And by (*), for every number k, if P (k) then P (k 0 ), so P (m0 ).
So, for all n, P (n).
¤
Lemma 3.1.2 The Principle of Complete Induction is a consequence of The
Principle of Mathematical Induction
Proof [Andrews] We want to show that if, for every number n, P (n) whenever
all number j < n are such that P (j), then P (n) for all n. So, we make the
following assumption:
(**) For all n, P (n) whenever all number j < n are such that P (j)
We must show that P (n) for all n. We prove this using The Principle of
Mathematical Induction.
Basis Step. n = 0. Since there are no j < 0, then by (**) it is trivial that
P (0).
Induction Step. The Induction hypothesis is that P (n). We want to show
P (n0 ). But, by (**), since P (n), then for all j < n, P (j). So, since P (n) and
for all j < n, P (j), it follows that, for all k < n0 , P (k). By (**), P (n0 ).
So, for all n, P (n). QED From these two lemmas, we derive
Lemma 3.1.3 The Principle of Mathematical Induction is equivalent to the
Principle of Complete Induction
Problems
3.1. The Principle of Mathematical Induction is a consequence of Course-ofValues Induction.
3.2. Course-of-Values Induction is a consequence of The Principle of Mathematical Induction.
3.3. The Principle of Complete Induction is a consequence of Course- ofValues Induction.
3.4. Course-of-Values Induction is a consequence of The Principle of Complete
Induction.
3.5. Suppose that the Basis step were eliminated from Course-of Values Induction to get:
Let P be a property of natural numbers. Suppose that:
If all natural numbers m ≤ n have the property P, then n0 has the property
P.
Then, every natural number n has the property P.
Show that this is equivalent to The Principle of Mathematical Induction.
3.1 Mathematical Induction
39
3.1.2 **Example: Parentheses
We take a very liberal attitude about what is to count as a property of numbers. We will express a property of numbers by means of an open sentence
with one free variable, assumed ranging over numbers. So, n is the number
of parentheses in a wff expresses a property of numbers, as does 2n is the
number of parentheses in a wff; also, n is the number of logical symbols in a
wff, n is the number of propositional symbols in a wff, n is the number of lines
in a proof of a wff, n is the length of a construction sequence for a given wff,
and so forth. Here, by way of example, we will carry out proofs by induction
on the number of parentheses.
[Kleene] Parentheses come in pairs, so that any formula will have 2n parentheses, n of them being left parentheses (, and n of them right parentheses ).
In fact, we might define such a set P of linear sequences of parentheses as follows. First, we suppose that we have a denumerable number of left parentheses,
(1 , (2 , . . . , (i , . . . and a denumerable list of right parentheses, )1 , )2 , . . . , )i , . . ..
Then we define our set P by induction:
Definition 3.1.1 (Kleene Parentheses)
(1) (i )i ∈ P
(2) If α ∈ P, then (i α)i ∈ P
(3) If α, β ∈ P, then αβ ∈ P.
(4) α ∈ P iff it can be generated by (1)-(3).
Definition 3.1.2 Two pairs of parentheses separate each other, if they occur
in the order (i (j )i )j , where the i’s identify either pair and the j’s the other,
and other parentheses may occur interspersed among the four shown.
Definition 3.1.3 A pairing of 2n parentheses is proper, if a left parenthesis
is always paired with a right parenthesis to the right of it, and if no two of the
pairs separate each other.
Now, we show that
Lemma 3.1.4 Every α ∈ P is properly paired
Proof by complete induction on n, where 2n is the number of parentheses in
α∈P
Induction Step. We assume that the theorem holds for all j < n. We have
to show that it holds for n. So, consider α ∈ P. If n = 1, then α = (i )i and so
is properly paired. If n > 1, then either α = (i β)i or else α = βγ. Each of β
and γ contain < n parentheses, so, by the induction hypothesis, β and γ are
both properly paired; so in either event, α must be properly paired.
So, for every n α is properly paired. QED
It is immediate that we can remove a pair of parentheses from a proper
pairing, and the remaining parentheses are properly paired. It is also obvious
that if (α) is properly paired, then α is properly paired.
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3 Mathematical Induction
Lemma 3.1.5 Every α ∈ P contains an innermost pair of parentheses, i.e.,
a pair which includes no other of the parentheses between them.
Proof by complete induction on n.
Induction Step. The induction hypothesis is that the Lemma holds for all
m < n.4 We must show that it holds for n.
Consider α ∈ P. Consider the leftmost parenthesis ‘(’. Either it and its mate
constitute an innermost pair of parenthesis, or there is a sequence of parentheses β between it and its mate which is a proper pairing of 2m parentheses,
m < n. By the induction hypothesis, β must have an innermost pair, and so
α must.
So, every such pairing must have an innermost pair. QED.
Lemma 3.1.6 A set of 2n parentheses admits at most one proper pairing.
Proof by complete induction on n.
Induction Step. The induction hypothesis is that for all m < n, a set of 2m
parentheses admits at most one proper pairing.
Now, by the previous lemma, a set of 2n parentheses has an innermost pair,
by the previous lemma. Remove this innermost pair and the remaining 2m
parentheses admit of at most one proper pairing, by the induction hypothesis.
So, the 2n admit of at most one proper pairing.
Lemma 3.1.7 Suppose α ∈ P contains β ∈ P as a subsequence. Then either
α = β or else α can be obtained from β by the constructions in the inductive
definition of P.
Proof by complete induction on the number n such that there are 2n parentheses in α.
Induction Step Assume the theorem holds for all j < n. We want to show
it holds for n.
If n = 1, then alpha and β have the same number of parentheses, so α = β. If
n > 0, then either α = (γ) or else α = γδ, the number 2j of parentheses in γ
and δ is such that j < n. So, by induction either β = γ or β = δ or else either
γ can be obtained from β or else δ can be obtained from β; in any case, α can
therefore be obtained from β. QED
3.1.3 The Principle of Induction on the Construction of a Wff
In this section, we want to prove
The Principle of Induction on the Construction of a Wff: Let F be a property of wffs, and let FA mean that A has the property F. Suppose that:
(1) Every propositional symbol in P has F;
(2) If A in P has F, then so does ∼ A
4
That is, P (n) is to mean ‘a proper pairing of 2n parentheses contains an innermost
pair’
3.1 Mathematical Induction
41
(3) If A and B in P have F, then so does (A ⊃ B)
Then every wff in P has F.
Proof
Let W be the set of wffs that have the property F. Then W contains all
propositional symbols (by (1)), and all complex wffs built up using our logical
connectives (by (2) and (3)). And since these just are the wffs (Recall the
definition of a wff), W contains every wff in P. QED.
We could, alternatively, have derived The Principle of Induction on the
Construction of a Wff using The Principle of Complete Induction. To do this,
we take n to be the length of a construction sequence for a wff.5 Induction
can be done for any property of numbers, and being the length of a wff is a
property of numbers. So, we proceed with complete induction.
Proof
To prove The Principle of Induction on the Construction of a Wff we will
make the assumption
(***) (1), (2), and (3)
and prove every wff has F. We will show that for every n, P (n), where P (n)
is Every wff having a construction sequence of length n has F. Since every wff
must have some a construction sequence of some length, this will suffice to
show that every wff has F. Our proof by complete induction then continues
as follows:
Induction Step. The hypothesis is that for every n, P (j) for all j < n.
Suppose A has a construction sequence of length n. Now, either A is a propositional symbol, in which case by (***) it has F, or else A is complex, and
either has the form ∼ B or the form (B ⊃ C), where B and C must each have
construction sequences < n. So, by the induction hypothesis B and C have F.
And by (***), therefore A has F. So, P (n).
So, for all n, P (n). QED
This means that all we need do to show that every wff A in P has a given
property, we need only show that every propositional symbol has the property
and every complex wff has the property.
3.1.4 **The Induction Principle
We have found that proof by induction rests on a more general principle
than we find using mathematical induction. If we look back to our Peano
Postulates, we find that we needed a starting element, or, better yet, a set of
starting elements from which we generated a set using a function. In the case
of the numbers, we started with 0 and generated the set of numbers using the
1-1 successor function. It’s being 1-1, however, was not needed for proof by
induction. In the case of the wffs, our starting set was the set of propositional
5
Note that a wff might have more than one construction sequence. This does not
affect the proof.
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3 Mathematical Induction
symbols, and then we generated the wffs using our wff-forming operations, ∼
and ⊃.
From Enderton (1972: 22ff). Consider an initial set B ⊆ U and a class
F of functions containing, for purposes of illustration, a binary operation
f : U × U → U and a singulary operation g : U → U .
Definition 3.1.4 A subset S of U is said to be closed under f and g iff
whenever elements x and y belong to S, then so do f (x, y) and g(x).
Definition 3.1.5 S ⊆ U is inductive iff B ⊆ S and S is closed under f and
g.
To say that a set is inductive, then, is to provide all but the last clause of an
inductive definition. The last clause can be filled in in two ways:
• From ‘the top down’, we say that the set defined, ‘C ∗ ’ is the smallest
inductive set. That is, ‘C ∗ ’ is the intersection of all the inductive subsets
of U .
• From ‘the bottom up’, we say that the set defined, ‘C∗ ’,will be the set of
elements of U which form the last element in a construction sequence.
Definition 3.1.6 A construction sequence is a finite sequence < x0 , . . . , xn >
of elements of U such that for each i ≤ n we have at least one of
xi ∈ B,
xi = f (xj , xk ) for some j < i, k < i,
xi = g(xj )
for some j < i
Let us verify that our two definitions are actually equivalent, i.e., that
C ∗ = C∗ .
Lemma 3.1.8 C ∗ ⊆ C∗
Proof. To show that C ∗ ⊆ C∗ we need only show that C∗ is inductive, i.e.,
that B ⊆ C∗ and C∗ is closed under the functions. Clearly B = C1 ⊆ C∗ .
If x and y are in C∗ , then we can concatenate their construction sequences
and append f (x, y) to obtain a construction sequence placing f (x, y) in C∗ .
Similarly, C∗ is closed under g. QED
Lemma 3.1.9 C∗ ⊆ C ∗
Proof. Finally, to show that C∗ ⊆ C ∗ we consider a point in C∗ and a construction sequence < x0 , . . . , xn > for it. By ordinary induction on i, we can
see that xi ∈ C ∗ , i ≤ n. First x0 ∈ B ⊆ C ∗ . for the inductive step we use the
fact that C ∗ is closed under the functions. Thus we conclude that
[
\
Cn = C∗ = C ∗ = {S: S is inductive}.
n
As a result, we get
Lemma 3.1.10 C ∗ = C∗
3.2 Unique Decomposition
43
Definition 3.1.7 The smallest set C ⊆ U such that B ⊆ C and which is
closed under the functions f and g is called the set generated from B by
f and g.
This enables us to state generally the following
Induction Principle Assume that C is the set generated from B by the functions f and g. If S is a subset of C which includes B and is closed under
the functions f and g. then S = C.
Proof. We are given that S ⊆ C. And since S is inductive, C ⊆ S since it is
the smallest such set. QED.
3.2 Unique Decomposition
3.2.1 Unique Readability Theorem
Our object in the present section is to show that each formula decomposes
uniquely. That means, in effect, that the functions used to generate the set
of wffs inductively is a 1-1 function. Put in a slightly different way, it means
that our use of parentheses assures us that each complex formula can be
built up from smaller formulas in one, and only one, way. Let us call the
various sentential connectives we use to create more complex wffs formula
building operations. Then we can say that if a complex wff has the form F 1 A,
there is no operation F 2 and wffs A, B such that F 1 A=F 2 (A, B); moreover,
if F 1 A = G 1 B, then F 1 = G 1 and A = B.6 So, the range of each formula
building operation is disjoint from every other formula building operation,
and each of the formula building operations is 1-1.
Each wff of P, then, must be exactly one of the following forms:
(i) propositional symbol
(ii) ∼ A
(iii) (A ⊃ B)
That is, every wff of P is either a propositional symbol or has a unique main
connective.
Further, given that a wff can be in only one of these categories, it can only
be in one of these categories in exactly one way. That is,
(i) If ∼ A =∼ B, then A = B
(ii) If (A ⊃ B) = (C ⊃ D), then A = C and B = D
The first lemma we need to establish is that every wff is going to have (at
least) one of the requisite forms.
Lemma 3.2.1 Every wff is going to be at least one of the following forms:
6
A similar claim is made for a complex wff built up using a binary operation.
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3 Mathematical Induction
(i) propositional symbol
(ii) ∼ A
(iii) (A ⊃ B)
Proof by complete induction on the length n of a construction sequence for the
wff D
Induction Step. The Induction Hypothesis is that the lemma holds for wffs
with construction sequence of length j < n. We must show it holds for wffs
with construction sequences of length n.
By the definition of construction sequence, each step in the sequence is either
a propositional symbol, or else it is the result of applying our wff building
operations to earlier lines, and so lines whose constructions sequences are of
length < n. Each of these, then, is going to be one of the appropriate forms,
and so the formula on step n will have to be arrived at using one of our formula
building operations. QED.
Definition 3.2.1 A formula is balanced if it contains the same number of
left and right parentheses
Lemma 3.2.2 Every propositional symbol is balanced.
Proof
Since a propositional symbol has no parentheses, it has the same number of
left and right parentheses.QED.
Lemma 3.2.3 Every wff is balanced.
Proof by induction on the construction of a wff
Every propositional symbol is balanced. And, if A and B are balanced, then
so are ∼ A and (A ⊃ B).
So, every wff is balanced. QED.
Lemma 3.2.4 No proper initial segment of a wff is a wff.
Proof by complete induction on the length of the construction of a wff
If our wff is a propositional symbol, it has no proper initial segment, so the
lemma follows trivially.
A proper initial segment of ∼ A, where A is a wff, will look like one of the
following:
∼
∼ A0 , where A0 is a proper initial segment of A
The first is clearly not a wff; and if ∼ A0 were wf, A0 would be wff, and by
the induction hypothesis, no proper initial segment of A is wf.
A proper initial segment of (A ⊃ B), where A and B are both wf, will look
like one of the following:
(
(A0 , where A0 is a proper initial segment of A
3.2 Unique Decomposition
45
(A
(A ⊃
(A ⊃ B0 , where B0 is a proper initial segment of B
(A ⊃ B
It is clear that none of these is balanced. QED.
We now show that the set of wffs is freely generated from the set of atomic
formulas by our two formula building operations.
Theorem 2 (Unique Readability Theorem). The two formula-building
operations, when restricted to the set of wffs,
(a) have ranges which are disjoint from each other and from the set of atomic
formulas, and
(b) are 1-1.
Proof of part (a). It is fairly obvious that no negation can also be a conditional.
¤
Proof of part (b). We have to show, now, that a wff is a conditional, in one
and only one way. So, we want to show that if A, B, C, D are all wffs and
(A ⊃ B) = (C ⊃ D)
then A = C and B = D. A cannot be a proper initial segment of C and C
cannot be a proper initial segment of A (Why?), so A = C. The next symbols
must be identical; and then, by similar reasoning, B = D. The case of negation
is handled in the same way.
¤
Exercise 3.2.1 Fill in the details of the Unique Readability Theorem
3.2.2 Finding the Main Connnective
[Enderton] Now, we want to describe an algorithm that, given a wff, will
produce its main connective.
(i) If the leftmost symbol is a propositional symbol, the wff is the propositional symbol and it has no main connective.
(ii) If the leftmost symbol is the negation symbol, then that is the main connective. The remainder is a wff and to find its main connective, go back
to step 1.
(iii) If the leftmost symbol is a left parentheses, scan the expression from the
left until first reaching a complete wff. Call this α. The next symbol must
be ⊃, and it is the main connective; the remainder of the wff will be a wff
β followed by a right parenthesis. To find the main connectives in α and
β, go back to step 1.
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3 Mathematical Induction
3.3 Some Theorems Involving Induction
3.3.1 Unique Valuation
An interpretation I is a map from each of the set of propositional symbols into
the set {T, F }, and the set of connectives {∼, ⊃} into f∼ and f⊃ respectively.
We want to show that any such interpretation I extends to a unique function
I 0 from the set of wffs into the set {T, F }. We define I 0 as follows:
If A is a propositional symbol, I 0 (A) = I(A)
If A and B are wffs, then
I 0 (∼A) = I(f∼ )I(A)
I 0 ((A ⊃ B)) = I(f⊃ )(I(A), I(B))
The proof is by Induction on the Construction of a Wff. We leave it as an
exercise to the reader.
3.3.2 The Interpolation Theorem
We proved the Interpolation Theorem informally back in Section 2.5; we now
want to prove it rigorously, using mathematical induction.
Claim (Interpolation Theorem). If |=P A ⊃ B, and A and B have at least one
propositional symbol in common, then there is a wff C of P containing only
propositional symbols common to both A and B such that |=P A ⊃ C and
|=P C ⊃ B.
Proof by Strong Mathematical Induction on n, the number of propositional
symbols that occur in A but not in B.
Basis Step: the case n = 0 is trivial: if every propositional symbol in A also
occurs in B, set C = A.
Induction Step: Assume the theorem holds for all k < n.
We need to show that it holds for n. Let p1 , ..., pn−1 , pn be the propositional
symbols occurring in A but not in B, p. Let q be a propositional symbol that
occurs in both A and B. (The theorem assures us that there is at least one.)
Since A ⊃ B is logically valid, it is T whether pn is T or F. We define the
following two wffs:
A1 = Aq⊃q
pn
A2 = A∼(q⊃q)
pn
Now, since
|=P A ⊃ B
then by our substitution principles,
|=P A1 ⊃ B
3.3 Some Theorems Involving Induction
47
|=P A2 ⊃ B
(Remember, pn only occurs in A, so substituting, e.g., q ⊃ q for pn in A
constitutes substituting q ⊃ q for pn in A ⊃ B.) Moreover, if A is T, then
either p is T or p is F, so, by our construction, either A1 is T (since p is
replaced by q ⊃ q) or A2 is T (since p is replaced by ∼ (q ⊃ q)). That is,
|=P A ⊃ (A1 ∨ A2 )
Furthermore, it is easily determined that
|=P [(A1 ⊃ B) ∧ (A2 ⊃ B)] ⊃ [(A1 ∨ A2 ) ⊃ B]
Modus Ponens enables us to infer
|=P (A1 ∨ A2 ) ⊃ B
The number of propositional symbols occurring in A1 ∨ A2 but not in B
is n − 1. Clearly, n − 1 < n, so by the induction hypothesis, there is some C
such that
|=P (A1 ∨ A2 ) ⊃ C
|=P C ⊃ B
Transitivity assures us that
|=P A ⊃ C
¤
3.3.3 Disjunctive Normal Form
Earlier we had argued informally that every wff is equivalent to one in Disjunctive Normal Form. Here we prove it rigorously.
Claim. Every wff A is equivalent to a wff D in Disjunctive Normal Form.
Proof by induction on the construction of a wff.
If n = 1, our wff is a propositional symbol and it is already in Disjunctive
Normal Form.
Suppose A is ∼ B. By hypothesis, B is equivalent to a wff in DNF: B will look
like this:
(b11 ∧ · · · ∧ b1n ) ∨ · · · ∨ (bj1 ∧ · · · ∧ bnm )
where each of the bij is a literal, so A will be
∼ [(b11 ∧ · · · ∧ b1n ) ∨ · · · ∨ (bj1 ∧ · · · ∧ bnm )]
DeMorgan’s Laws permit us to run the negation sign in:
∼ (b11 ∧ · · · ∧ b1n ) ∧ · · · ∧ ∼ (bj1 ∧ · · · ∧ bnm )
48
3 Mathematical Induction
And then again:
(∼ b11 ∨ · · · ∨ ∼ b1n ) ∧ · · · ∧ ∼ (bj1 ∨ · · · ∨ ∼ bnm )
Assume all double-negations have been tended to. Then, cross-multiplying
gives us:
(∼ b11 ∧ · · · ∧ ∼ bj1 ) ∨ · · · ∨ (∼ b1n ∧ · · · ∧ ∼ bnm )
And we are in DNF. Call this wff D.
Suppose A is (B ⊃ C), and by hypothesis, each of B and C are equivalent to
wffs in DNF. We know this is equivalent to ∼ B ∨ C We have just shown that
∼ B can be put into DNF if B is in DNF, and since C can be put into DNF,
∼ B ∨ C can be put into DNF since the disjunction of two wffs in DNF is in
DNF. QED.