Math 3040 Spring 2011

The Propositional Calculus†

Contents

1. Truth-value 22. Elementary logical operations and truth-values 22.1. Identity 32.2. Negation 32.3. Conjunction 32.4. Disjunction 32.5. Implication 32.6. Truth-value dependence 43. Mod 2 arithmetic and truth-values 54. Truth tables for the elementary logical operations 55. General logical operations and logical expressions 66. Truth tables for general logical operations 77. Logical equivalence 98. Rules of Inference and Proofs 108.1. Rules of Inference 128.2. Formal methods of proof 139. Some practical tips for proving theorems 15

Any stepwise procedure consists of what we might call “treads” and “risers”. In mathe-matical proofs, the treads are mathematical statements, and the risers involve logical rulesand rules of inference. This chapter deals with rules of logic and inference and how thesemay be used in the construction and understanding of proofs.

We begin with a simplified version of symbolic logic, the propositional calculus, in whichso-called quantification does not occur. The propositional calculus analyzes the truth re-lationships between compound statements and their subsidiary parts. A later chapter, onthe predicate calculus, expands on this by introducing quantification.

Symbolic logic is sometimes called formal or syntactic, in the sense that, after the basicrules have been established, symbolic logic focuses exclusively on the form of statementsand not on their meaning. The idea is to find universal rules that apply to all complexstatements, depending only on their form.

Needless to say, mathematics, just as any other systematic field of study, is very muchconcerned with the meaning of its assertions. However, a careful analysis of these assertionsreveals that certain formal, logical relationships occur again and again. Therefore, it isprofitable to study these formal relationships in their own right as tools or short cuts forconstructing and understanding mathematical statements.

Profitability aside, when the basic tools of symbolic logic are combined with the conceptsof set theory, as we do later, the resulting mix is powerful enough to generate virtually all

† c©January 12, 2009

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of what we mean by mathematics. Later chapters will substantiate this claim by showinghow to develop many of the notions and theorems of analysis, algebra, and geometry in thisway.

1. Truth-value

As defined in the Glossary, a statement is a meaningful assertion that is either true orfalse. It is sometimes convenient to express the truth or falsity of a statement as though itwere a certain numerical value assigned to the statement. We call this its truth-value. Moreprecisely, if P is a statement that is true, we say that its truth-value is 1 and express thisas tv(P ) = 1; if P is false, we say that its truth value is 0 and write tv(P ) = 0.

2. Elementary logical operations and truth-values

Suppose we are considering two statements, which we call P and Q for short. TheGlossary describes four ways of using logical connectives to construct new statements fromthese:

negation: notP , written symbolically as ¬P ;conjunction: P and Q, written symbolically as P ∧Q;disjunction: P or Q, written symbolically as P ∨Q;implication: P implies Q, written symbolically as P ⇒ Q;

We adjoin to these one more operation, which is “trivial,” in the sense that it doesnothing: it is the identity operation, which, given, say, P , just produces P again. This isnot interesting in and of itself, but it is convenient to have when discussing more generaloperations later.

These constructions are called elementary logical operations. The symbolic expressionon the right corresponding to each operation is called an elementary logical expression. Itgives a convenient shorthand for describing the operation and for building more complexoperations. The statements P and Q in the expressions are called atomic statements, ormore simply, atoms of the expression, since they are not resolved more finely into subsidiarystatements. The entire statement formed in this way from the atoms is sometimes called acompound statement.

It is remarkable that all of what we call deductive reasoning (without quantification)consists of what we can construct by repeating or combining the above simple operations.Before going into how such combination works, let us first look at the relationship betweenthe operations and truth values.

We are interested in how the elementary logical operations behave with respect to truth-values; i.e., how the truth values of

P, ¬P, P ∧ Q, P ∨ Q, P ⇒ Q

relate to the truth-value of the atom P and the truth value of the atom Q. Note that weare not simply playing a game with symbols here. We are interested in using the symbolismwe introduced to model what actually happens when we, as human beings, reason aboutcertain aspects of the world described by various statements. Therefore, truth-values shouldbe assigned in such a way as to reflect these well-known patterns in reasoning. That is, weshould determine the appropriate assignment of truth values for each elementary operationby referring to its meaning.

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2.1. Identity. This operation does nothing to the given statement P , so it obviously doestruth

valuesof

compoundstatements

not affect the truth-value.

2.2. Negation. By definition, the negation ¬P of the statement P asserts the contraryof P . This means that when tv(P ) = 1, then tv(¬P ) = 0, and when tv(P ) = 0, thentv(¬P ) = 1.

2.3. Conjunction. Since P ∧Q asserts both P and Q, we have tv(P ∧Q) = 1 when bothtv(P ) = 1 and tv(Q) = 1, but tv(P ∧Q) = 0 otherwise. That is, tv(P ∧Q) = 0 wheneverat least one of tv(P ) = 0 or tv(Q) = 0 and only then.

2.4. Disjunction. By definition, P ∨ Q affirms at least one of the statements P or Q.Therefore, tv(P ∨Q) = 1 whenever at least one of tv(P ) = 1 or tv(Q) = 1; tv(P ∨Q) = 0otherwise, i.e., when both tv(P ) = 0 and tv(Q) = 0.

2.5. Implication. The truth-values produced by the operations of identity, negation, con-junction and disjunction flow in a straightforward way from their meanings in ordinaryEnglish. The case of implication is somewhat different.

When we say P implies Q in ordinary English, we usually refer to a situation in which Qis causally or inferentially connected to P . For example, when we say “If it is sunny tomor-row, then Jane will go swimming,” we have in mind a fairly complex set of circumstancesconnecting the weather to Jane’s activities.

The mathematical operation P ⇒ Q, however, does not delve into such connections. Yes,the definition is motivated by the ordinary usage of the word “implication.” But technically,the definition is expressed only in terms of truth-values. The implication P ⇒ Q is deemedto be true unless it is falsified. And it is falsified (i.e., false) precisely when P is true andQ is false. Notice that when P is false, P ⇒ Q cannot be falsified, so it is true in this case.We sometimes say in this case that P ⇒ Q is vacuously true.

We now give a few examples of P ⇒ Q (or “if P , then Q”).(a) Let n denote an integer. If n is even and is a perfect square, then n is divisible by 4.(b) Suppose you are given a right triangle. If one of the angles of the right triangle is

45◦, then the triangle is isosceles.(c) Consider a real number x. If x2 + 1 = 0, then x is negative.(d) Let R denote a square. If R has side of length 3, then the area of R equals 6.

In example (a), if n = 16, then both P and Q are true. But if n = 7, both are false.However, in either case, P ⇒ Q is true because it is not falsified. Indeed, there is no valueof n that falsifies P ⇒ Q in this example.

Example (b) is similar. Either both P and Q are true (which happens when one of theangles is 45◦) or both are false (when no angle is 45◦). In either situation, P ⇒ Q is notfalsified, so it’s true.

In example (c), no value of x satisfies P . That is, it is false whatever x is considered.Therefore, P ⇒ Q is not falsified for any x, and so it’s true for all x.

In example (d), P is false precisely when R has a side of length other than 3, so P ⇒ Qis vacuously true in all these cases. However, when the side has length 3, then P is trueand Q is false, so P ⇒ Q is then false.

To summarize, we have the following rules about the truth-values of implication:

tv(P ⇒ Q) = 1 when either tv(P ) = 0 or when tv(Q) = 1;

tv(P ⇒ Q) = 0 when tv(P ) = 1 and tv(Q) = 0.

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There is a more convenient way of summarizing this rule by using truth tables, as we explainshortly.

A number of ordinary-English constructions are used as synonyms for “P implies Q”: “ifP , then Q,” “P only if Q,” “P is sufficient for Q,” “Q is necessary for P , “Q is a consequenceof P ,” “Q follows from P .” The statement P is often called the hypothesis or antecedent ofthe implication, and the statement Q is often called the conclusion or consequence.

Exercise 1. Convince yourself, by using specific examples of statements for P and Q, thatthe statements just mentioned are truly synonyms for “P implies Q.”

Exercise 2. Some statements do not appear to be implications, but they can be reformu-lated so as to be implications. For example, the statement “The square of a real number isnon-negative” can be reformulated as “If x is a real number , then x2 ≥ 0. ” Reformulatethe following statements as implications:

• Every differentiable function is continuous.• All right angles are equal.• A matrix with a zero eigenvalue is not invertible.

Exercise 3. Identify each of the following statements as a non-trivial compound statement,indicating the atoms and the logical operation that are used. In case the statement is animplication, indicate which atom is the hypothesis and which is the conclusion. (Note: thestatement is not necessarily true.)

• A necessary condition for a positive integer n to be a prime is that it be odd.• Base angles of an isosceles triangle are equal.• All that glitters is not gold.• Both 97 and 87 are prime.• At least one of the integers 71 and 73 is prime.

2.6. Truth-value dependence.The elementary logical connectives are modeled after our real-world usage of these terms,

and we use the real-world meanings of the connectives to justify our assignment of truthvalues to the resulting compound statements. However, we are not concerned with theparticular meanings of the atomic statements. Our assignment of truth values has thefeature that a compound statement S formed from atoms P and Q via an elementarylogical operation has truth-value tv(S) depending only on tv(P ) and tv(Q) and not on anyfurther aspects of the meaning of P or of Q. To put it another way that will be familiarto students who have studied calculus, tv(S) is a function of tv(P ) and tv(Q). In any case,for the purposes of looking at truth-values, we are liberated from any complex analysis ofmeanings (i.e., from semantics) and can focus entirely on the formal, syntactic way thatcompound statements are constructed from their atoms.

These logical constructions, which we look at more carefully later, have an algebraic flavorto them, since both propositional calculus and traditional algebra involve the manipulationof atomic symbols and connectives. This similarity is not accidental. One of the foundingworks of symbolic logic is George Boole’s 1848 treatise, “The Laws of Thought,” in whichhe developed symbolic logic in an algebraic form. The modern descendant of his work is

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known as a Boolean algebra; it plays an important role in computer design. The relevant“number” system for such algebra is the field of two elements, or what is sometimes called‘mod two arithmetic.’

3. Mod 2 arithmetic and truth-values

The symbolism used for the elementary logical operations is suggestive of algebraic no-tation, and, indeed, we shall soon discuss how to combine the operations in an algebraicmanner. However, we can already use fairly standard algebra when working with the truth-values of statements and the elementary logical operations. These truth-values are numbers0 and 1, and there is a well-known ‘mod two’ arithmetic that we shall apply to these.

We all have learned that even and odd integers when added or multiplied satisfy certainrules, such as odd + odd = even, odd + even = odd, odd × odd = odd, odd × even = even,etc. If we use the number 0 to represent even numbers and the number 1 to represent oddnumbers, then these rules can be displayed in an addition table and a multiplication tableas follows:

+ 0 10 0 11 1 0

and× 0 10 0 01 0 1

These tables define the operations of mod 2 addition and multiplication To simplifynotation, we use the usual convention for multiplication and replace the × sign by simplejuxtaposition. Notice that since addition and multiplication are binary operations (thatis, they operate on two numbers at a time), we need to put parentheses around a sumor product when using it in a more complicated expression, just as in ordinary algebra.Sometimes these parentheses may be dropped when there is no possibility of error.

We now use mod 2 arithmetic to express the truth values of elementary logical opera-tions algebraically. Let P and Q be any given statements, and let tv(P ) be denoted by xand tv(Q) by y.

Exercise 4. Verify the following equalities in mod 2 arithmetic.(a) tv(¬P ) = 1 + x.

algebraicrepresen-tationoflogicalexpressions

(b) tv(P ∧Q) = xy.(c) tv(P ∨Q) = x + y + xy.(d) tv(P ⇒ Q) = 1 + x + xy.

Note that. for each elementary compound statement S involving the atoms P and Q, theabove give explicit formulas for tv(S) in terms of tv(P ) and tv(Q), substantiating what wesaid earlier.

4. Truth tables for the elementary logical operations

From now on, let us consider some set S of statements that is closed under the elementarylogical operations. That is, whenever P and Q are statements in S, then ¬P , P ∧Q, P ∨Q,and P ⇒ Q are also statements in S.

For each such elementary logical expression, its truth-values can be completely describedby a small table, which lists all the possible combinations of truth-values of the atoms, andthen, next to each combination, gives the corresponding truth-value of the expression. Sucha table is called a truth table:

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P Q P ∧Q0 0 00 1 01 0 01 1 1

P Q P ∨Q0 0 00 1 11 0 11 1 1

P Q P ⇒ Q0 0 10 1 11 0 01 1 1

P ¬P0 11 0

P P0 01 1

Figure 1. The truth tables of the elementary logical operations

The information in these tables just repeats the information given above in 2.1 — 2.5,but it displays it in a more convenient format. Notice that the number of rows in each truthtable depends only on the number of atoms in the elementary logical expression. If thereis only one atom, as in negation and identity, the table has two rows plus the header row;if there are two atoms, as in the three other cases, there are four rows plus the header row.The format of truth tables extends easily to more general logical operations, as we describenext.

5. General logical operations and logical expressions

In using the elementary logical connectives, we can start with one or two statements inS and produce another statement in S, and so it is possible to use the end product again in

compoundstatements a similar way. That is we can apply elementary logical operations to statements repeatedly

to build more and more complex expressions, each representing a compound statement.For a random example, suppose that we are given statements L,M, P, Q in S. We may

first form, say, L∧M and P ∨Q, then form ¬(L∧M) and P ∨(¬(L∧M)), and then, finally,

(P ∨Q) ⇒ (P ∨ (¬(L ∧M))).

We have inserted parentheses to indicate that what they enclose is to be regarded as de-noting a statement. This avoids possible notational ambiguity when there are many logicalconnectives floating around in the expression. Since there is no ambiguity in the case of theentire statement, we do not put parentheses around the entire expression.

The resulting expression describes a statement in S. It is built by applying the elementarylogical operations to some of the statements L,M,P, Q to get new statements, then usingthese and the original statements and logical connectives to get further statements, etc.,finally, applying an elementary operation to the two penultimate statements to get the finalexpression. In this particular case, the last operation performed is an implication appliedto the penultimate statements P ∨ Q and P ∨ (¬(L ∧M)). The expression we obtain forthe end result is a convenient representation of how it is constructed.

Here is another very simple example that deserves special attention because it is used sooften: namely, consider the expression

(P ⇒ Q) ∧ (Q ⇒ P ),

which, of course says that P implies Q and Q implies P . Because of its frequent use, itlogical

equivalence receives a special symbol, just like one of the elementary operations: namely, we write

P ⇐⇒ Q.

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Most typically, we express this in English as “P if and only if Q,” but we also sometimesuse as synonyms “P is equivalent to Q,” “P is logically equivalent to Q or “P is necessaryand sufficient for Q.” Mathematicians sometimes use the shorthand notation “iff” to standfor “if and only if.”

Exercise 5. Convince yourself that the above synonyms really do say the same thing aboutthe relationship between P and Q as “P if and only if Q.”

In general, a logical operation is defined to be the application of a sequence of elementarylogical operations to statements P, Q, R, . . . of S, as in the two examples above. The result iscalled a logical expression, and P, Q,R, . . . are called its atoms, just as before. This descrip-tion can be made more precise and formal, but that is not needed for our purposes. We maydenote such a general logical operation or expression by such symbols as L(X1, X2, . . . , Xr)or perhaps G(P1, P2, . . . , Pr), or F (P, Q,R, S), where X1, X2, P1, P2, P, Q etc., denotethe atoms in the expressions. When we do not care to be specific about naming the atoms,we may write the expressions simply as L, G, or F .

Given logical expressions F and G, we may choose to go further and use them to buildother more complex logical expressions by using elementary logical operations on them. Forexample, we might want to construct F and then apply negation to it. We can denote thissimply as ¬F . Or, we might want to construct F and G and then form the conjunctionof the two. This can be conveniently denoted by F ∧G. Therefore, using this convention,we can apply all of the elementary logical operations to F and G, obtaining new logicalexpressions

F, ¬F, F ∧G, F ∨G, F ⇒ G.

We also may apply the non-elementary logical operation ⇐⇒ to form F ⇐⇒ G, withthe same understanding as above as to what this means.

6. Truth tables for general logical operations

Since a general logical expression is built by successively applying elementary operations,its truth-value depends only on the truth-values of the individual atoms. Therefore, just asfor elementary logical expressions, a general logical expression has a truth table. This isa rectangular array with initial columns (starting on the left side) headed by the atoms ofthe expression, and the last column (on the right) headed by the expression itself. Theremay be intermediate columns to represent subsidiary terms of the expression. Each rowcorresponds to a specific assignment of truth values to the atoms. So, if there are r atoms,there must be 2r rows to accommodate all the possible combinations of such truth-valueassignments.

Exercise 6. Check the validity of this last assertion for the value r = 4. Can you explainwhy it is valid in general?

In each row the atoms are assigned some set of truth values, and then a truth value isfilled in for every subsequent column according to the rules established for the elementaryoperations.

As an example of such a truth table, let us consider a the following logical expressionwith three atoms, say, F = F (P, Q,R) = (P ∧R) ⇒ (Q ∨R).

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P Q R P ∧R Q ∨R F0 0 0 0 0 10 0 1 0 1 10 1 0 0 1 10 1 1 0 1 11 0 0 0 0 11 0 1 1 1 11 1 0 0 1 11 1 1 1 1 1

Figure 2. The truth table of (P ∧R) ⇒ (Q ∨R)

Since the number of atoms is three, the truth table of F has eight rows. We have insertedcolumns corresponding to the subsidiary terms P ∧R and Q ∨R.

This example of a truth table is fairly typical in most respects,except that its last column consists entirely of 1’s. This does not happen in general, as

can be seen by looking at the truth tables for the elementary logical operations. When thisdoes happen, however, it means that no matter what the truth-values of the atoms, theend result is always true. Therefore, such an operation is what we called in the Glossary a

tautologiestautology. To summarize, a tautology is a logical expression such that the last column in itstruth table consists entirely of 1’s.

As noted in the Glossary, the opposite of a tautology is called a contradiction. This cancontradic-

tions be described as a logical expression whose truth table has only 0’s in its last column.Obviously, if a logical expression F is a tautology, then ¬F is a contradiction, and,

conversely, if F is a contradiction, then ¬F is a tautology.

Exercise 7. Verify that each of the following logical expressions is a tautology by computingits truth table:

(a) P ∨ ¬P .(b) ¬(P ∧ ¬P ).(c)

(P ⇒ ¬(¬P )

) ∧ (¬(¬P ) ⇒ P).

(d) P ⇒ (P ∨Q).(e) (P ∧ (P ⇒ Q)) ⇒ Q.(f) (P ⇒ Q) ⇒ (¬Q ⇒ ¬P ).(g) (¬Q ⇒ ¬P ) ⇒ (P ⇒ Q)(h)

((P ⇒ Q) ∧ (Q ⇒ R)

) ⇒ (P ⇒ R).

Exercise 8. Construct the truth table for P ⇐⇒ Q, including columns for the subsidiarystatements P ⇒ Q and Q ⇒ P .

If you do this exercise correctly, you will see that P ⇐⇒ Q is true precisely when P andQ have the same truth-values, and it is false otherwise.

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Exercise 9. Construct truth tables for the following logical expressions:(a) (P ∨Q) ⇒ P .(b) P ⇒ (P ∧Q).(c) (¬P ∨Q) ∧ (¬R ∨ P ).

Exercise 10. (a) Display the truth tables for A = P ∧Q, B = ¬P ∧Q, C = P ∧ ¬Q,and D = ¬P ∧¬Q. Then compute the truth tables for A∨B, C∨D, and B∨C ∨D.

(b) Suppose you are given a column 4-tuple of 0’s and 1’s. Explain how to constructan expression with atoms P and Q whose truth table has that as its last column.(Hint: If the 4-tuple has only 0’s, construct some contradiction involving P and Q.If the 4-tuple has at least one 1, explain how a suitable disjunction of A, B, C, andD from part (a) will do the trick.)

(c) Consider truth tables for expressions involving exactly two atoms (i.e., four rows plusreducedtruthtables

the row of headers). Call such a table reduced if it has no intermediate columns, i.e.,it has only columns for the atoms and for the entire expression. Show that there areat most 16 reduced truth tables for logical expressions involving two atoms.

(d) Use (b) to show that there are at least 16 reduced truth tables for logical expressionsinvolving two atoms.

(e) What would you guess the number to be for expressions with three atoms? Or natoms? How would you prove that these are indeed the correct numbers? (Don’ttry to give a proof. Just explain what you would need to verify.)

7. Logical equivalence

Now let F and G be any logical expressions involving the same number of atoms, sayr of them, each atom ranging over all the statements in S. We shall say that F andG are logically equivalent, provided that, for all possible statements, P1, P2, . . . , Pr in S,F (P1, P2, . . . , Pr) has the same truth-value as G(P1, P2, . . . , Pr). According to the commentafter Exercise 8, we may formulate this as follows: F and G are logically equivalent providedthat the compound expression F ⇐⇒ G is a tautology, i.e., the last column of the truthtable for F ⇐⇒ G consists entirely of 1’s. We may abbreviate this by simply writingF ⇐⇒ G.

This definition has the following obvious consequences for any logical expressions F,G,and H:

(a) F ⇐⇒ F .(b) If F ⇐⇒ G, then G ⇐⇒ F .(c) If F ⇐⇒ G and G ⇐⇒ H, then F ⇐⇒ H.

Each of these assertions may easily be checked by simply looking at the truth-values of theexpressions F,G, H for any choice of truth-values of atomic statements P1, P2, . . . , Pr.

The following exercise presents properties of logical equivalence which show that it re-spects the elementary logical operations.

Exercise 11. Suppose that F ⇐⇒ F ′ and G ⇐⇒ G′.(a) Verify that

propertiesoflogicalequivalence

¬F ⇐⇒ ¬F ′, (F ∧G) ⇐⇒ (F ′ ∧G′), (F ∨G) ⇐⇒ (F ′ ∨G′),

(F ⇒ G) ⇐⇒ (F ′ ⇒ G′).(b) Verify that (F ⇐⇒ G) ⇐⇒ (F ′ ⇐⇒ G′).

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(c) Suppose that L is any logical expression involving two atoms. Can you show thatL(F, G) ⇐⇒ L(F ′, G′)? (Hint: Show that L(F, G) has the same reduced truth-table as L(F ′, G′).)

This exercise suggests the general fact that we may always substitute logical expressionsfor logically equivalent ones without affecting truth-values. We shall be using this generalfact frequently.

It remains to give some useful examples of logical equivalences. The following exerciselists eight of the most commonly used equivalences.

Exercise 12. Verify the following logical equivalences:commonly

usedequiva-lences

(a) P ∧ (Q ∨R) ⇐⇒ (P ∧Q) ∨ (P ∧R)(b) P ∨ (Q ∧R) ⇐⇒ (P ∨Q) ∧ (P ∨R)(c) ¬(¬P ) ⇐⇒ P(d) ¬(P ∧Q) ⇐⇒ ¬P ∨ ¬Q(e) ¬(P ∨Q) ⇐⇒ ¬P ∧ ¬Q(f) P ⇒ Q ⇐⇒ (¬P ) ∨Q(g) P ⇒ Q ⇐⇒ ¬Q ⇒ ¬P(h) ¬(P ⇒ Q) ⇐⇒ P ∧ ¬Q.

Notice that equivalences (a) and (b) resemble the distributive law for multiplication andaddition in ordinary arithmetic.

The implication on the right in item (g) is known as the contrapositive of the implicationthe

contra-positive

P ⇒ Q, as described in the Glossary. This logical equivalence leads to a very useful methodfor proving implications, as we discuss later. Logical equivalence (f) also involves implica-tion. It shows that implication may be expressed in terms of negation and disjunction.

Equivalences (d) and (e) are known as DeMorgan’s Laws. They show how negation affectsDeMorgan’s

Laws disjunction and conjunction, which is often useful in proofs. The following exercise, asksthe student to apply De Morgan’s laws to slightly more complicated expressions.

Exercise 13. Derive the following logical equivalence from the logical equivalences in Ex-ercise 12, giving your reasons for each step.

¬ (P ∨ (Q ∧R)

) ⇐⇒ (¬P ∧ ¬Q) ∨ (¬P ∧ ¬R).

8. Rules of Inference and Proofs

The Glossary describes a proof of a mathematical statement P as a valid chain of rea-soning that begins with statements known to be true and concludes with P . More precise

deductiveproofs terminology would probably call this a description of a deductive proof, since it presents

a stepwise process in which each step involves a deduction or inference. Indeed this isthe notion of proof that we shall be focusing on throughout this course. It is, however,worth pointing out that there is a more general concept of proof that does play a role inmathematics.

In its most general sense, a proof can be defined simply as a convincing argument, oreven better, as a convincing demonstration. Logical steps and rules of inference are not anecessary part of this definition. With this notion of proof, the demonstration may appearholistically, perhaps visually, and it is often at least as convincing as any deductively linked

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chain of statements. We give one example of this, which is a well-known proof of thePythagorean Theorem.

Recall that the Pythagorean Theorem states that in a right triangle with sides of lengths aand b and hypotenuse of length c, the quantities a, b and c satisfy the relation a2 + b2 = c2.There are many, many proofs of this theorem. Euclid’s proof, among many others, is adeductive proof, since it is essentially a compilation of deductive steps that begin withEuclid’s axioms for geometry. But there are also holistic proofs, that is, proofs whichdemonstrate the result more or less directly by means of a picture. The following pairof diagrams comprise the picture that provides perhaps the most direct and convincingexample of this.

a

b

b a

c a

b

c

SS

SS

SS

SS

ZZ

ZZ

ZZ

ZZ

SS

SS

SS

SS

½½

½½

½½

½½

½½

½½

½½

½½

SS

SS

SS

SS

The two large squares pictured are assumed to be congruent, so that they have the samearea. In each square, there are four right triangles, all congruent but assembled differently.The viewer is invited to draw the obvious conclusion that the remaining regions in eachlarge square have equal areas. In the right-hand picture, the remaining region is a squarewith side of length c—hence, area c2— whereas in the other picture, the remaining region iscomprised of two non-overlapping squares of sides a and b, respectively—hence, with totalarea a2 + b2. So, as Agatha Christie’s Hercule Poirot might say, “Voila, Hastings, my littlegrey cells perceive that the Theoreme of Pythagoras is correct.”

Now the careful reader may notice that there are a few picky questions left open by thisargument. For example, Why are all the small triangles congruent? For another example,Why is the quadrilateral on the right with side designated “c” a square? Specifically, Whyare all of its angles right angles?

Of course, it is not very hard to answer these questions. But if one does so carefully andreasonably completely, then one is simply re-proving some of the introductory theorems inEuclid’s Elements! That is, one is transforming or reducing the picture-proof to a deductiveproof. To my knowledge, there is no proof, holistic or otherwise, that cannot be reduced toa deductive proof in roughly this manner. It follows that deductive proofs can be taken asthe universal proof procedure, provided we are principally interested in using the method ofproof as a validating tool !

Indeed, that is what we shall do in this course. The above “picture-proof” shows thatother kinds of proof are possible and may reveal geometric relationships or stimulate otherintuition in the way some deductive proofs would not. This is their great strength.

To be sure, the discovery of a deductive proof often also requires ingenuity or inspiration.Further, the careful construction of such a proof, or, alternatively, the careful reading ofsuch a proof, may reveal relationships and intricacies not immediately apparent, thereby

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providing a deeper understanding of the proposition being proved for anyone who goesthrough this process. So, there is no clear winner between these competing paradigms.Rather, we have to remain open to the advantages of each.

Two additional points about picture proofs or other possible “holistic” or “intuitive”proofs. One is that, almost by definition, they resist systematization. It is very difficult toteach how to have a stroke of inspiration. The other point is that pictures can be misleading,particularly if they are complex or require that various choices be made. A well-known false“proof” that all triangles are isosceles depends on just such a misleading picture.

Because the construction of a deductive proof proceeds by breaking a problem down intosmaller problems, and then perhaps these, in turn, into yet smaller ones — cf., the “divideand conquer” rule of computer programming — the deductive approach has the advantageof reducing some of the amount of high-level ingenuity required while at the same timereducing the probability of error. When we follow the deductive paradigm, then, there is achance that useful mathematical results can be correctly proved by all of us.

What we describe in this section are techniques for beginning a deductive proof of somestatement P according to the form of P as a compound statement. This often reduces theproblem of constructing a proof to smaller problems which can then be attacked in the sameway. However, the specific choices and steps that you use to “fill in” this architecture willstill require your active imagination.

To begin, we discuss the basic dynamic principles in proofs: rules of inference. Wethen discuss different possible logical structures for proofs—the “architecture”— and giveexamples. The rules of inference and logical structures will all be based on our discussionof the propositional calculus. In a later chapter, we adjoin to these some additional formalprinciples coming from the predicate calculus.

In addition, we give a number of practical tips for doing proofs in the appendix thatfollows this section. These will also be discussed in class. Versions of some of them can befound in Solow’s book, “How to Read and Do Proofs.”

In subsequent parts of these notes, we shall encounter additional tools of proof, includingthe important method of induction as well as various counting arguments.

8.1. Rules of Inference. As defined in the Glossary, a rule of inference is what allowsus to proceed correctly from one set of statements to another statement. It provides the

modusponens dynamic that allows us to prove theorems. There is one basic rule of inference in logic from

which any other may be derived. It has a Latin name, modus ponens, because it was firstused by the ancient Greeks. :-). We shall first say informally what modus ponens is, andthen we discuss how it ties in with logical operations.

Modus ponens tells us that if S and T are two statements such that S and S ⇒ T aretrue, then we may deduce that T is true. This follows from the meaning of implication, aswe discussed earlier, because in the presence of a true S, a false T provides precisely thecriterion for S ⇒ T to be false. And this last is contrary to assumption. Therefore, the twotrue statements S and S ⇒ T lead to the true statement T . To summarize:

S true and (S ⇒ T ) true yield T true.The modus ponens rule also may be viewed as simply our interpretation of the tautology

(P ∧ (P ⇒ Q)) ⇒ Q ( cf. Exercise 7 (e)).Modus ponens can also be used with logical expressions, say F and G, both built from

atomic statements P1, P2, . . .. Suppose that some choices of these atomic statements are suchthat the corresponding expressions F (P1, P2, . . .) and F (P1, P2, . . .) ⇒ G(P1, P2, . . .) are

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true. Then, modus ponens allows us to conclude that G(P1, P2, . . .) is true. The reasoningfor this is the same as before.

Modus ponens can be used to derive other rules of inference. We give one examplehere. Suppose we know that T is false and that S ⇒ T is true. We may then use modusponens to prove that S is false. For, were S true, modus ponens would imply that T

proofbycontra-positive

is true, which it isn’t. So S is false. This is the basis for so-called proof by contrapos-itive, which we discuss further below. So, modus ponens implies the rule of inference:

T false and (S ⇒ T ) true yield S false.This rule also has a Latin name: modus tolens. However, we shall not use this name

modustolensbecause modus tolens and other such rules can be derived from modus ponens. Instead, we

usually say something like “by an application of modus ponens.”

Exercise 14. Show that the following rule can be derived from modus ponens: SupposeP, Q and R are statements such that P ⇒ Q is true and Q ⇒ R is true. Then, the statementP ⇒ R is true. (Hint: Use Exercise 7(h).)

8.2. Formal methods of proof. Every proof in mathematics may be viewed as the deriva-tion of some conclusion B from some hypothesis A. Sometimes A remains tacit, as part ofthe background, but often it is given explicitly. The task is to prove that the implicationA ⇒ B is true. Once this is done, the validity of A produces the desired validity of B bymodus ponens.

But how do we prove A ⇒ B? The detailed answer to this, of course, will depend onthe particulars of A and B. In some cases, we may use a tautology to transform A ⇒ Binto a logically equivalent statement that is easier to prove. We describe these and someexamples below. In other cases, the implication A ⇒ B does not admit a convenient formalreduction to some more easily proved statement. We describe some approaches to dealingwith such implications in the appendix.

In any case, either after reduction to another statement or not, one is left with one ormore implications to prove. We’ll still call one of these A ⇒ B for now. For each suchimplication, one basic step is always used: namely, we assume that A is true, and then usethat information, together with any other knowledge we have that is appropriate to themathematical context, to derive the statement B. Notice that to proceed, we do not needto know whether A is actually true or not; we simply assume the truth. This approach isjustified in precisely the same way that we justified the truth-values in the truth table forimplication. For, to assert A ⇒ B means only that B holds whenever A does. So, if wecan verify that B is true under the assumption that A is true, we have eliminated the onlypossibility for the implication to fail.

We now give a number of formal techniques for beginning the proof of A ⇒ B. Theyare called “formal” techniques, because they are classified according to the particular formthat A or B takes and are independent of the meanings of A and B. In each case, A ⇒ Bwill be replaced by a logically equivalent statement according to some tautology. Solow’sbook uses such a classification and provides many examples. In the examples we present, weshall be assuming facts from earlier mathematics courses, such as basic facts from geometry,algebra, and so on. Solow’s book provides a convenient, short list of some of these in hisChapter 3, p.24.

A couple of additional methods will be introduced after we introduce quantification (inThe Predicate Calculus).

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proof by contradiction: Assume A and ¬B and derive a contradiction C. Thismethod is based on the logical equivalence

(A ⇒ B) ⇐⇒ ((A ∧ ¬B) ⇒ C),

where C is any (conveniently chosen) contradiction. Since a self-contradictory state-ment such as C may appropriately be considered absurd and the method involvesdeducing (or reducing to) C, this method is often called reductio ad absurdum, an-other example of a Latin phrase used to describe a Greek method.1 This logicalequivalence can be verified by checking truth tables.

proof by contrapositive: Assume ¬B and derive ¬A. This method is based on thelogical equivalence

(A ⇒ B) ⇐⇒ (¬B ⇒ ¬A).

Often this procedure is phrased slightly differently. Instead of assuming that ¬B istrue and deriving that ¬A is true, we assume that B is false and deduce that A isfalse. Clearly these two formulations amount to the same thing, according to ourunderstanding of what “true” and “false” mean.

disjunctive hypothesis method or either/or method, type I: This method ap-plies when the statement A is a disjunction, say A = E ∨ F . In this situation, youmust do two proofs: namely, you must prove both E ⇒ B and F ⇒ B. Thus,separately, assume E and derive B and also assume F and derive B. This methodis based on the logical equivalence

(E ∨ F ⇒ B) ⇐⇒ ((E ⇒ B) ∧ (F ⇒ B)).

disjunctive conclusion method or either/or method, type II: This method ap-plies when the statement B is a disjunction, say B = E ∨ F . In this case you havea choice of one of two possible methods. Either assume both A and ¬E and thenderive F ; or assume both A and ¬F and then derive E. This method is based onthe two logical equivalences

(A ⇒ (E ∨ F )) ⇐⇒ ((A ∧ ¬E) ⇒ F )

and(A ⇒ (E ∨ F )) ⇐⇒ ((A ∧ ¬F ) ⇒ E).

Exercise 15. Verify each of the logical equivalences used in the two “either/or” methodsabove by computing truth tables for the expressions on each side of the equivalence.

More generally, any logical expression that is logically equivalent to A ⇒ B can be usedto obtain a method of proof of A ⇒ B. The ones just described above are perhaps the mostcommon, but they are not the only ones. Again, the particulars of the statements A and Bwill suggest which method to use.

We now give a few examples. We won’t always give complete proofs here, only thetranslations of the above methods into concrete terms.

1The student may wonder about this phenomenon of Latin names of Greek methods. The reasons for this arefairly simple. The methods were known through the preservation of manuscripts that had been translatedinto Latin. The names were devised in the Middle Ages when Latin was well known to the educated butGreek was not.

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Examples of the methods above

• For proof by contradiction, we use a simple example involving positive integersn. Let A be the statement “n2 is odd,” and let B be the statement “n is odd.”Then, to prove A ⇒ B by contradiction, we assume that both n2 is odd and n is notodd (i.e., it’s even) and derive a contradiction. We leave this last step to the reader.

• For proof by contrapositive, we could use the previous example and try to provethat ¬B ⇒ ¬A, which translates to: n even implies n2 even. The reader should beable to check this immediately. This example shows that proof by contradiction andproof by contrapositive are very similar. Often it’s simply a matter of taste to useone or the other. We’ll discuss this further in class.

• For the disjunctive hypothesis method or either/or method, type I, supposethat T denotes either an equilateral triangle or a square, and consider the followingstatement: If T has perimeter of length P , then its area is < P 2/12. First noticethat, although the hypothesis does not look like a disjunction, it really is, becauseT is either an equilateral triangle of perimeter P or a square of perimeter P . So,the implication is proved by first verifying that < P 2/12 when T is an equilateraltriangle of perimeter of length P and then verifying the inequality when T is a squarewith perimeter of length P . (Or do these in the reverse order.) We do not work outthe geometric proof here but may discuss this in class.

This example suggests that this method of proof can be thought of as dealingwith two separate cases; in each case separately, the conclusion must be verified.From this viewpoint, there is nothing special about the fact that we are dealing withtwo cases; there could be three or even more. In fact the hypothesis A could be adisjunction E ∨ F ∨ G ∨ . . . of many statements, thus requiring separate proofs ofeach of the implications E ⇒ B, F ⇒ B, G ⇒ B, etc. This could then be calledthe method of proof by cases. The logical equivalence given above to justify themethod for two cases easily extends to as many cases as we like. But, to repeat: Onlywhen every one of the cases is proved do you have a proof of the entire implication.

• For the disjunctive conclusion method or either/or method, type II, let Abe the statement “x is a real number satisfying x2 = 1, and let B be the disjunction“x = 1 or x = −1.” Then, the disjunctive conclusion method would have us assumeA and, say, x 6= 1, and then use these to prove x = −1. We’ll give a complete proofin this case:

(1) Assuming A and performing the algebraic step of transposition, we get x2−1 =0.

(2) Factoring this expression, we get (x− 1)(x + 1) = 0(3) Using the assumption x 6= 1, we conclude that x− 1 6= 0.(4) Dividing equation (2) by x− 1, we get x + 1 = 0, from which x = −1 follows.

This concludes the proof.

9. Some practical tips for proving theorems

A foreword: Every student attending this course already knows an enormous amountof mathematics. This includes both logical and problem-solving skills. The challenge is tobring this knowledge to bear effectively. The formal techniques developed in this chapter

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should be helpful in designing a proof of a given implication. But there are some additional,simple tips that can also be useful. Your aim should be both to come up with an idea forthe proof and to communicate this proof efficiently, clearly and correctly.

Here, I should give an important caveat: There is no procedure or finite list of proceduresfor proving every theorem, just as there is no super method for solving every problem. Tobe sure, there are computer proof systems that can prove certain limited kinds of theorems.But the fact (due to Kurt Godel) is that, for every such system, there always exist truestatements that can be formulated in the system but are not provable in the system. Twomore practical considerations are: (a) It has not been demonstrated that most theorems ofinterest to mathematicians are amenable to such systems, and (b) The programs encodingsuch theorem-proving systems are hardly suitable for human use.

So, we should not expect there to be a single “Method” whereby every mathematicaltheorem can be proved or problem solved. The best we can hope for is to find a tool kitthat will enable us to tackle the problem of proof on a case by case basis.

(a) Preparing for the proof. This should always be your first step (or collection ofsteps).

(i) Make sure you understand the context of the problem. For example, is thissomething you are showing to a friend? Something you are writing on a home-workpaper? A proof on an exam, with instructions accompanying the prob-lem? And so on. What background knowledge are you allowed to assume?How detailed should your writeup be?

(ii) Then, read the proposition that is to be proved carefully and write it in theform of an implication. Sometimes this won’t be necessary because it is alreadywritten that way. However, often the implication will be couched in morecomplicated language and it will be important to sort out exactly what thehypothesis is and what the conclusion is. Especially if you are having troublegetting started on a proof, writing out the implication explicitly can be veryhelpful.

(iii) Decide whether the implication is in one of the forms described in §8.2 and, ifit is, transform it as described there. You will then have ended up with oneor more implications that require proof.

(iv) Say one of them is A ⇒ B. Be sure you understand what A is saying andwhat B is saying. The idea is to assume that A is true and then deduce thatB is true.

(v) See the comments below in “What to do when you are stuck.” They may behelpful in kick-starting your proof.

(b) Synthesis versus analysis. The Greek mathematician Pappus identified two ap-proaches to proving an implication, which he called “synthesis” and “analysis.”Solow calls these the “forward” and “backward” methods.

(i) In synthesis, you use everything you know about A, together with whateveryour are allowed to assume in the given context, and you derive B.

I’ll give an example of this now. I’ll do this in two parts. The first goesthrough my thought processes as I am trying to work out a proof. This sort of

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“thinking-through” should appear on your “scratch-paper”, but it should notappear in your finished proof or in an exam paper unless it’s explicitly askedfor! The second part will be the finished proof.

Illustrative problem 1: Prove that the product of twoodd integers is odd.

As part of my proof preparation, I’ll start by rewriting the state-ment as an implication and using symbols. My re-write is: Ifm and n are odd integers, then mn is an odd integer.So, here, A is “m and n are odd integers,” and B is “mn is anodd integer.”

Next I look at background and context: I have to assume that Iam allowed to use some prior knowledge about integers...shortof my knowledge that mn must be odd when both m and n are,of course. So, I assume that I can use the fact that an integeris odd if and only if it can be written in the form 2k + 1, forsome integer k. Then I have “m = 2k + 1 and n = 2j + 1.”This is essentially statement A.

Next, I must derive B, and I observe that B involves multipli-cation. I conclude that I am allowed to use prior knowledgeabout arithmetic. Therefore, to derive B, I compute: mn =(2k +1)(2j +1) = 2(2kj +k + j)+1. So, mn is an odd integer,which is statement B.

I’ll now give what I feel is a good, finished proof of the implication:

(1) Let m and n be odd integers. We must provethat mn is odd.

(2) Since m and n are odd, there are integers k andj such that m = 2k + 1 and n = 2j + 1.

(3) Therefore, using standard arithmetic, we get mn =(2k+1)(2j+1) = 2(2kj+k+j)+1, which impliesthat mn is odd.

This completes the proof.

This proof, although probably briefer than most, can serve as a model foryour finished proofs. Please note the following: (a) There is no display ofthe thoughts that led to the proof. (b) At the same time, each step is ac-companied by a brief statement explaining and, if necessary, justifying whatis being done. When you are in doubt as to how much of a reason you shouldgive, remember that it is better to err on the side of saying a little bit toomuch. (c) In steps (1) and (2), I describe the variables m, n, k, j that Iam using. I don’t expect you (the reader) to know what these are withoutmy telling you. (d) In step (1), after introducing the variables m and n,I say explicitly what I am going to prove in terms of the new notation. It is

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important to give the reader this information so that he/she knows what tolook for. And it is also a simple, routine way to get started with writing andconstructing the proof (see comments below in xxx).

Note that in synthesis (i.e., in the “forward”) method, we just plow straightahead, using whatever we know or are allowed to assume. This is fine in somesituations, but sometimes it is unclear what direction to go or what data shouldbe used . Indeed, sometimes A itself may consist of all of prior mathematics,and how do you sort that out? For these kinds of issues, the method of analysiscan be very useful.

(ii) In analysis, we are still trying to prove A ⇒ B. But instead of using informa-tion about A to plow ahead and derive B directly, we try to find a statementB′ that has two properties: (I) We can prove B′ ⇒ B; (II) A ⇒ B′ seems tobe easier to prove than A ⇒ B.

Admittedly, item (II) is a bit vague. But suppose you can find a B′ thatsatisfies (I) and, better than just showing that it satisfies (II), you actually canprove A ⇒ B′! Then, you could use the tautology

(A ⇒ B′ ∧ B′ ⇒ B) ⇒ (A ⇒ B),

together with modus ponens to conclude that A ⇒ B. You are done!

Okay, but suppose you can find a B′ satisfying (I) and (II), but you can’t proveA ⇒ B′? What then? Well, you have made the problem slightly easier. Sonow repeat this backward step. Find another statement B′′ such that (I) Youcan prove B′′ ⇒ B′; (II) A ⇒ B′′ seems easier to prove than A ⇒ B′. If youcan actually prove A ⇒ B′′, then the tautology gives you A ⇒ B′, and then,as before, you get A ⇒ B.

Clearly, if even this second effort doesn’t suffice, you can try to repeat theprocess. Although there is no guarantee, in many cases you will be successfulafter a number of steps. Schematically, you will have found a sequence of newstatements B′, B′′, . . . , B∗ such that you can prove sequence of implications

A ⇒ B∗ ⇒ . . . ⇒ B′′ ⇒ B′ ⇒ B.

And, as we have seen, such a sequence implies that A ⇒ B, as required.

This procedure is said to work backwards from B.

As a practical matter, you rarely will need to do more than a couple of stepsin this process.

But how do you find a new statement B′ to start the analysis method? Onegood way to proceed is to follow the advice of Solow (pp. 10 -12) and ask aso-called key question about B. An answer to this question should generatethe desired new statement B′. If that is sufficient, fine, but if not, ask a keyquestion about B′, etc. Solow gives a number of examples of this process.

I’ll now give one example here. Again, I first go through my reasoning, andthen I write a finished proof.

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Illustrative problem 2: Prove that

limx→∞(x− 1)(x− 2)(x− 3)(x− 4)(x− 5)(x− 6)/x7 = 0.

Reasoning: (a) I first notice that the problem involves a limit inwhich the variable goes to infinity. So I’ll assume the context ofcalculus, which is where I first studied such limits. Writing thestatement as an implication is not obvious, since the problemgives only the concluding statement B. So, this is an examplein which A is tacit...that is, A consists of all the basic definitionsand general properties in calculus and algebra pertaining topolynomial expressions and limits of real variables. So, I canjust use these and omit writing a particular A.

(b) Next I should come up with a key question. Looking at B,it seems reasonable to ask “How do I prove that the limit offunction values equals zero?” This is a general question. Oneanswer would be to invoke the definition of limit. That seemspretty complicated. Instead, I try to remember some facts fromcalculus about limits that will prove useful.

One fact that suggests itself to me is the so-called “sandwichlemma”: If f(x), g(x), h(x) are real-valued functions such thatf(x) ≤ g(x) ≤ h(x), for all x in some interval containing a, andlimx→a f(x) = L = limx→a h(x), then limx→a g(x) = L. So apossible answer to the key question might be, “Use the sand-wich lemma to imply that the limit of function values is zero.”Unfortunately, this doesn’t immediately give a statement B′.So I then ask, “How can I choose f(x), g(x), h(x) so that thesandwich lemma will imply B?”

(c) Of course, g(x) should just be given by the expression

g(x) = (x− 1)(x− 2)(x− 3)(x− 4)(x− 5)(x− 6)/x7.

No problem. But what about f(x) and h(x)? Well, I want bothof these to approach 0, so I could just take f(x) to be constantly0. The only trouble with this is that g(x) is not ≥ 0 for allx. However, I am only interested in values of x in an intervalcontaining ∞, which means that I may restrict x to be largerthan some constant: say x > 6. For such x, we do get g(x) ≥ 0,i.e., letting f(x) be the constantly zero function, 0 = f(x) ≤g(x). So that works. But what about h(x)? Well, for anyx > 6, the positive numbers x−1, x−2, x−3, x−4, x−5, x−6are each ≤ x. So g(x) ≤ x6/x7 = 1/x. Therefore, we canchoose h(x) = 1/x, defined for all x > 0. We now have thefunctions f(x), g(x), h(x) appearing in the sandwich lemma,and they satisfy f(x) ≤ g(x) ≤ h(x), as needed.

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(d) Notice that

limx→∞h(x) = lim

x→∞(1/x) = 0.

We may conclude this from basic facts in calculus (all of whichwe have agreed to label as A). So, letting B′ be the statementlimx→∞ h(x) = 0, we have A ⇒ B′. The proof of B′ ⇒ B iscontained in the discussion above. So, by the scheme describedearlier, this implies A ⇒ B.

Therefore, we have worked backwards from the given statement Bto a simpler one B′, which was easy to prove (using facts fromcalculus). The process of finding B′ was such that we coulddeduce B from B′.

Finished proof: We shall prove the stated limit,assuming basic facts about limits from calculus.

(1) Since we are interested only in the limit as x →∞, we may restrict to x > 6.

(2)For x > 6, 0 ≤ (x− 1)(x− 2)(x− 3)(x− 4)(x−5)(x− 6)/x7 ≤ x6/x7 = 1/x.

(3)limx→∞(1/x) = 0, by results of calculus.

(4) Using (2) and (3), together with the “sandwichlemma” from calculus, we may conclude that

limx→∞(x− 1)(x− 2)(x− 3)(x− 4)(x− 5)(x− 6)/x7 = 0,

which is what was to be proved.

One brief comment. Numbering important steps in a lengthy proof or number-ing equations in a long computation can be very useful for reference purposes.Research papers or textbooks in mathematics do this all the time. Generally,it is not necessary to number the steps in short proofs as I have done in thetwo proofs above. However, it does have the advantage of clearly indicatingexactly what the steps in the proof are, and it may help avoid the insertionof extraneous material. So, I’d recommend your numbering your steps duringthe next three or four weeks, even for the short proofs that you produce.

(c) What to do when you are stuck. Mathematicians often get stuck when theytry to solve problems or construct proofs. In fact, I’d guess that 90% of a mathe-matician’s time is spent trying to get “unstuck” with a problem he or she is workingon. This should not be surprising, because if the problem were easier than that,either it’s probably not very interesting or someone else would have solved it sometime ago. In any case, because of this, mathematicians develop various routines forgetting unstuck. What follows are some of the things that I do. Sometimes thesework. Sometimes not. But they’re worth trying.

(i) This one was mentioned before. Write out the implication carefully, under-standing exactly what you may assume (i.e., A, together with background

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material) and what you need to prove. Introduce only as much notation asnecessary for writing A and B.

(ii) Try to transform the implication to another form, using one of the tautologiesdescribed before, or any other tautology from the propositional calculus thatgives a statement implying A ⇒ B. This sometimes does work: a new for-mulation may be easier to work with. But I wouldn’t put too much time intothis.

(iii) Draw a picture or a schematic diagram illustrating the problem. If the problemis in geometry, usually this means drawing a picture of the geometric objectsabout which something must be proved. But even if the problem is in algebra,or calculus, or set theory, etc., a diagram indicating the relations betweenthe objects in the problem can be very useful. Examples of this will appearrepeatedly in class or in exercises.

(iv) Sometimes you will be asked to prove a general fact involving expressions invariables that range over large sets (e.g., like the set of real numbers). (Thiswill be made more precise in The Predicate Calculus. See especially §14.3of that chapter.) It can be very useful to choose a variety of values of thevariables, evaluate the expressions and check whether the desired result holdsfor those choices of values. Generally this does not constitute a proof!You are just looking at special examples. However, these examples mightbe suggestive of why the general fact is true: they might suggest a proof.Alternatively, they might suggest that the purported “fact” is incorrect: i.e,they might be counterexamples. The incorrectness of the purported generalfact can be formulated as the truth of the negation. Therefore, the so-calledcounterexamples you have computed will constitute a proof of this negation.

In any case, it is always important to look at special examples when you arestuck in working on a proof, or when you are just starting the proof.

(d) A few traps to avoid.

(i) Do not assume what you are trying to prove. If you want to prove A ⇒ B,you may assume A, but you may not assume B! Sometimes it is easy toget lost as to what is the conclusion that you are aiming to prove. This isespecially true when you start using transformations of the problem, such asthe contrapositive or proof by contradiction, and it also happens when studentsuse the method of analysis (i.e., the “backwards” method). Keep foremost inyour mind that B is the conclusion you are aiming to prove; it is not somethingyou may assume.

(ii) Sometimes you will be asked to prove a general fact and you simply check oneor two special cases. Most of the time this is not good enough. The proof youare asked to give is supposed to hold in general, not for just the special casesyou choose. We discuss this issue further in The Predicate Calculus, and we’llgo over examples in class.

(iii) Are you using all the information that you’ve been given? If not, you might stillbe in good shape because the unused information might have been superfluous!

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Usually not, though. Usually, one of two things will happen: (a) You willdevise a proof that is incorrect, or (b) You will not be able to proceed at allwith the proof because you are missing necessary facts.

I have seen case (a) a number of times. Students (and even professional mathe-maticians) devise intricate (and usually slightly confusing) proofs. After somethought, it appears that the proof applies to an absurdly wide variety of situa-tions, situations where it obviously cannot be correct. And this is because theproof is not constrained by some of the data given in the problem. Exampleswill be given in class.

Case (b) is easier to cope with, because then you are really stuck insteadof being misled by a false proof. Often, when you make use of the missinginformation, the problem becomes much easier.

The moral of the story is that you should always be sure that you have used allthe information given in the problem.

Exercise 16. Each of the following assertions is a true statement. For each one, start bydescribing your reasoning as you construct a proof, and then give a finished proof. Explicitly,if you change the form of the assertion, indicate which formal method you use. Then, stateif you use synthesis or analysis, and, if you use analysis, describe your key questions.

(a) The area of an isosceles right triangle equals one fourth the square of the hypotenuse.(b) If n is a perfect square, then n has the form 4k or 4k + 1, for some integer k.(c) Suppose that p(x) = ax2 + bx + c is a real polynomial such that b2 − 4ac < 0 and

c > 0. Then, p(x) > 0 for all real values of x.(d) Suppose that p(x) is a real polynomial satisfying p(x) = p(−x), for all real values

of x. Then, no odd powers x2k+1 appear in p(x). (Note: We say that a power xn

“appears in a polynomial” if its coefficient is non-zero.)