3 Mathematical Induction

Reading: Metalogic Part II, 23,27

Contents 3.1 Mathematical Induction ...... 35 3.1.1 Peano’s Axioms ...... 35 3.1.2 **Example: Parentheses ...... 39 3.1.3 The Principle of Induction on the Construction ofaWﬀ...... 40 3.1.4 **The Induction Principle ...... 41 3.2 Unique Decomposition ...... 43 3.2.1 Unique Readability Theorem ...... 43 3.2.2 Finding the Main Connnective ...... 45 3.3 Some Theorems Involving Induction ...... 46 3.3.1 Unique Valuation ...... 46 3.3.2 The Interpolation Theorem ...... 46 3.3.3 Disjunctive Normal Form ...... 47

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,... 36 3 Mathematical Induction

This informal characterization of the set of natural numbers N, is formally characterized by Peano’s Postulates1

P1 0 ∈ N. 0 P2 If n ∈ N, then there is a unique n ∈ N. 0 0 P3 For all m, n ∈ N, if m = n , then m = n. 0 P4 For each n ∈ N, n 6= 0. 0 P5 If M ⊆ N such that 0 ∈ M and m ∈ M whenever m ∈ M, then M = N.

0 P1 is straightforward. P2 tells us that is a function, so that if x = y, then 0 0 0 x = y .P3 tells us that the function 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 eﬁect, 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 These are actually due to Dedekind 2 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, 0 by P2. And, clearly n 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) identiﬂes 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. 38 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 (k0), 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-of- Values Induction.

3.2. Course-of-Values Induction is a consequence of The Principle of Mathe- matical Induction.

3.3. The Principle of Complete Induction is a consequence of Course- of- Values 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 In- duction 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 deflne 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 deflne our set P by induction: Definition 3.1.1 (Kleene Parentheses) (1) (i)i ∈ P (2) If α ∈ P, then (iα)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 = (iβ)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. 40 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 Wﬁ: Let F be a property of wﬁs, 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 wfi in P has F. Proof Let W be the set of wfis that have the property F. Then W contains all propositional symbols (by (1)), and all complex wfis built up using our logical connectives (by (2) and (3)). And since these just are the wfis (Recall the deflnition of a wfi), W contains every wfi 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 wfi.5 Induction can be done for any property of numbers, and being the length of a wfi 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 wfi 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 wfi must have some a construction sequence of some length, this will su–ce to show that every wfi 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 wfi A in P has a given property, we need only show that every propositional symbol has the property and every complex wfi has the property.

3.1.4 **The Induction Principle

We have found that proof by induction rests on a more general principle than we flnd using mathematical induction. If we look back to our Peano Postulates, we flnd 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 wfis, 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. 42 3 Mathematical Induction symbols, and then we generated the wfis using our wfi-forming operations, ∼ and ⊃. From Enderton (1972: 22fi). 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 deflnition. The last clause can be fllled in in two ways: • From ‘the top down’, we say that the set deflned, ‘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 deflned, ‘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 deﬂnitions 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 efiect, that the functions used to generate the set of wfis inductively is a 1-1 function. Put in a slightly difierent 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 wfis formula building operations. Then we can say that if a complex wfi has the form F 1A, there is no operation F 2 and wfis A, B such that F 1A=F 2(A, B); moreover, if F 1A = G1B, then F 1 = G1 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 wfi of P, then, must be exactly one of the following forms: (i) propositional symbol (ii) ∼ A (iii) (A ⊃ B) That is, every wfi of P is either a propositional symbol or has a unique main connective. Further, given that a wfi 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 flrst lemma we need to establish is that every wfi 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. 44 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 wfis with construction sequence of length j < n. We must show it holds for wfis with construction sequences of length n. By the deflnition of construction sequence, each step in the sequence is either a propositional symbol, or else it is the result of applying our wfi 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 wfi 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 wfi 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 wfi, will look like one of the following: ∼ ∼ A0, where A0 is a proper initial segment of A

The flrst is clearly not a wfi; and if ∼ A0 were wf, A0 would be wfi, 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 wfis 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 wfi is a conditional, in one and only one way. So, we want to show that if A, B, C, D are all wfis 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 wfi, will produce its main connective. (i) If the leftmost symbol is a propositional symbol, the wfi 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 wfi and to flnd its main connective, go back to step 1. (iii) If the leftmost symbol is a left parentheses, scan the expression from the left until flrst reaching a complete wfi. Call this α. The next symbol must be ⊃, and it is the main connective; the remainder of the wfi will be a wfi β followed by a right parenthesis. To flnd the main connectives in α and β, go back to step 1. 46 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 I0 from the set of wfis into the set {T,F }. We deflne I0 as follows: If A is a propositional symbol, I0(A) = I(A) If A and B are wfis, then 0 I (∼A) = I(f∼)I(A) 0 I ((A ⊃ B)) = I(f⊃)(I(A), I(B)) The proof is by Induction on the Construction of a Wfi. 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 wfi 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 deflne the following two wfis:

q⊃q A1 = Apn ∼(q⊃q) A2 = Apn 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 wfi is equivalent to one in Dis- junctive Normal Form. Here we prove it rigorously. Claim. Every wfi A is equivalent to a wfi D in Disjunctive Normal Form. Proof by induction on the construction of a wff. If n = 1, our wfi is a propositional symbol and it is already in Disjunctive Normal Form. Suppose A is ∼ B. By hypothesis, B is equivalent to a wfi 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 wfi D. Suppose A is (B ⊃ C), and by hypothesis, each of B and C are equivalent to wfis 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 wfis in DNF is in DNF. QED.