Lectrure Material> Introduction to Logic Lecture 9 Brian Weatherson, Department of Philosophy

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Introduction to Logic Lecture 9 Brian Weatherson, Department of Philosophy

September 30, 2009

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Informal Proofs

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What are Proofs

A proof that an argument is valid is: 1

A sequence of statements;

2

The first statements are the premises of the argument;

3

The last statement is the conclusion of the argument;

4

Each statement follows directly from earlier statements;

5

And we say exactly how each statement follows from earlier statements.

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Formal and Informal Proofs

We write formal proofs out using the Fitch notation.

We write informal proofs in English, either in sentences or in bullet points.

In these slides I’ll often use bullet points, or even numbered bullet points, to make it easier to see what I’m referring to.

There are a lot of different ways of doing formal proofs.

We are using the Fitch system because it looks the most like reasoning in English, especially when the English version is in numbered sentences.

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Follows Directly

One of our conditions on a proof was: Each statement follows directly from earlier statements.

But what does it mean to say that something ‘follows directly’ ?

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Follows Directly

In practice, what follows directly depends a lot on what your audience is.

Some things that might be obvious in an engineering lab might need to be spelled out in a law court, and vice versa.

In intro logic, at least at the start, we spell out everything as clearly as we can.

As we move through the course, we’ll get more and more liberal about what steps we allow.

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Starting Steps

But even in intro logic, there are a few steps that are too obvious to need more justification. These include: From P ∧ Q, infer P.

From P ∧ Q, infer Q.

From P and Q, infer P ∧ Q. From P, infer P ∨ Q.

From Q, infer P ∨ Q.

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Proof by Cases

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Suppositions

It’s time to complicate a little our picture of what goes on in a proof.

Consider this situation: A, B and C are baseball teams.

Each of them has won 90 games this season, and no other team has won more games.

C has played all its games for the season.

A and B have one game left, and they will be playing each other.

In baseball, every game must end with one or other team winning.

How might we prove that at least one team will end up with more wins than C ?

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Here is one way the proof might go.

1

A and B are playing against each other.

2

So one of A and B will win that game.

3

Suppose A wins the game.

4

Then A will have 91 wins.

5

So A will have one more win than C .

6

So a team has more wins than C .

7

Suppose B wins the game.

8

Then B will have 91 wins.

9

So B will have one more win than C .

10

So a team has more wins than C .

11

So, one way or the other, a team will end up with more wins than C .

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Here is one way the proof might go.

1

A and B are playing against each other.

2

So one of A and B will win that game.

3

Suppose A wins the game.

4

Then A will have 91 wins.

5

So A will have one more win than C .

6

So a team has more wins than C .

7

Suppose B wins the game.

8

Then B will have 91 wins.

9

So B will have one more win than C .

10

So a team has more wins than C .

11

So, one way or the other, a team will end up with more wins than C .

That’s a good proof, but what is going on at lines 3 and 7?

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Suppositions

The answer is that we are supposing something.

A supposition is not something that follows from the premises.

So it is not something that we are claiming must be true.

But it is something that helps to structure the proof.

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Suppositions

The tricky thing about suppositions is knowing when to discharge them.

To discharge a supposition means to say something like this: From our original premises, plus supposition X , conclusion Y follows.

So Z must follow from the original premises alone.

As a rule, the first of those steps is easy, there are few restrictions on it, but the second step is hard.

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Proof By Cases

The proof we gave above was one kind of case where we can use a supposition. It followed the following form.

We have earlier derived that X ∨ Y is true.

We supposed X , and derived Z .

Then we dropped the supposition that X , instead supposed Y , and again derived Z .

From all that we concluded that Z must be true.

As we might put it slightly more informally, either X or Y must be true, and either way, Z is going to follow. We’ll illustrate this with a couple more examples.

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Souvenir Example

Suppose Max’s favorite basketball team is the Chicago Bulls and favorite football team is the Denver Broncos. Max’s father John is returning from Indianapolis to San Francisco on United Airlines, and promises that he will buy Max a souvenir from one of his favorite teams on the way. Explain John’s reasoning, appealing to the annoying fact that all United flights between Indianapolis and San Francisco stop in either Denver or Chicago.

Make explicit the role proof by cases plays in this reasoning.

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1

John is flying from Indianapolis to San Francisco on United.

2

All United flights from Indianapolis to San Francisco stop in Chicago or Denver.

3

So John’s flight will stop in Chicago or Denver.

4

Suppose it stops in Chicago.

5

Then he’ll buy a Chicago Bulls souvenir.

6

The Chicago Bulls are Max’s favorite basketball team.

7

So John will buy a souvenir from one Max’s favorite teams.

8

Suppose instead it stops in Denver.

9

Then he’ll buy a Denver Broncos souvenir.

10

The Denver Broncos are Max’s favorite football team.

11

So John will buy a souvenir from one Max’s favorite teams.

12

Either way, John will buy a souvenir from one Max’s favorite teams.

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Burglary Example

Suppose the police are investigating a burglary and discover the following facts. All the doors to the house were bolted from the inside and show no sign of forced entry. In fact, the only possible ways in and out of the house were a small bathroom window on the first floor that was left open and an unlocked bedroom window on the second floor. On the basis of this, the detectives rule out a well-known burglar, Julius, who weighs two hundred and fifty pounds and is arthritic. Explain their reasoning.

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1

The burglar came in through the bathroom window or the bedroom window.

2

Suppose the burglar came in through the bathroom window.

3

The bathroom window is small.

4

So the burglar is small.

5

Julius weighs 250 pounds, which is not small.

6

So Julius is not the burglar.

7

Suppose instead that the burglar came in through the bedroom window.

8

The bedroom window is on the second floor.

9

So the burglar was able to climb to the second floor.

10

Julius is arthritic.

11

An arthritic burglar can’t climb to the second floor.

12

So, Julius is not the burglar.

13

Either way, Julius is not the burglar.

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Proof by Cases and Excluded Middle

In all the cases that we’ve looked at so far, we’ve had to be given the disjunction X ∨ Y .

But in some cases that disjunction comes for free.

Those are cases where Y = ¬X .

For any X , we know X is either true or false.

So either X is true or ¬X is true.

So X ∨ ¬X is true.

So if we can show something follows from both X and ¬X , we’ll have shown that it must be true.

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Mathematics Example In the textbook they work through a proof of a surprising mathematical result that uses proof by cases, starting with an instance of X ∨ ¬X . n An irrational number is one that cannot be written as m for integers n, m.

Question: Are there irrational numbers b, c such that b c is rational?

The answer turns out to be yes.

The obvious way to answer the question goes by finding the particular b and c.

The proof they give doesn’t do that; rather, it shows that some such b and c must exist.

The proof uses two important mathematical facts that you might have forgotten from high-school algebra.

First, for all x, y , z ∶ (x y )z = x yz . √ Second, 2 is irrational.

Now, onto the proof.

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√ √2 Either 2 is rational or it is irrational.

2 Assume first that it is rational. √ 3 Then letting b = c = 2, we see there are b, c such that b c is rational. √ √2 4 Assume instead that 2 is irrational. √ √2 √2 5 Now consider ( 2 ) . √ √2×√2 6 By the theorem we saw above, this must equal 2 . √ √ √ 7 But by the definition of , 2 × 2 = 2. √ √2 √2 √ 2 8 So ( 2 ) = 2 √ √ 2 9 And again by the definition of , 2 = 2. √ √2 √2 10 So ( 2 ) = 2. √ √2 √2 11 So ( 2 ) is rational. √ √2 √ 12 So, setting b = 2 and c = 2, there are b, c such that b c is rational.

13 So either way, there are b, c such that b c is rational.

1

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Proof By Contradiction

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Indirect Proof Here is a common way to prove ¬P across different fields of study.

1

First, we assume P.

2

From that we derive some absurd result.

3

We conclude that our initial assumption was wrong.

4

That is, we conclude ¬P.

What counts as absurd is different in different fields.

In engineering, it might be the existence of a perpetual motion machine.

In finance, it might be the existence of a foolproof strategy for beating the market.

In logic and mathematics, it is an explicit contradiction.

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Contradiction

By an explicit contradiction, we mean a pair of statements of the form X and ¬X .

Whenever we have such a pair of statements, we say that we’ve derived a contradiction.

And, in general, we’ll assume at that point that our last assumption was false.

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Euclid on Prime Numbers This style of proof goes back at least thousands of years.

Over 2300 years ago, Euclid used it (or reported on others using it) to prove that there must be infinitely many prime numbers.

The proof uses a number of mathematical facts.

First, if there are finitely many integers with a certain property, then there is a largest integer with that property.

Second, if x = kn + 1, for any integers x, k and any integer n > 1, then x is not divisible by n. The reason for this is that kn is divisible by n, and the next number divisible by n is kn + n, which is bigger than kn + 1.

Third, we write n! for the product of the first n integers. So 6! is 1 × 2 × 3 × 4 × 5 × 6 = 720.

Now here’s Euclid’s proof.

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Assume that there are finitely many prime numbers.

Then there is a largest prime number. Call it n.

3 Now consider the number n! + 1.

4 If there is some prime p < n! + 1 that divides n! + 1, it is either less than or equal to n, or greater than n.

5 For any prime less than or equal to n, that number divides n!, so it can’t divide n! + 1.

6 And by hypothesis, there are no primes greater than n, so no prime greater than n can divide n! + 1.

7 So no prime p < n! + 1 divides n! + 1.

8 That means that n! + 1.

9 Since n! + 1 > n, that means that there is a prime greater than n.

10 So we’ve derived a contradiction: there are no primes greater than n, and there is a prime, namely n! + 1, that is greater than n.

11 So our initial assumption, that there are finitely many primes, must be false.

12 So there are infinitely many primes.

1 2

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Here is a simpler example.

Assume n2 is odd. Prove n is odd.

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Here is a simpler example.

Assume n2 is odd. Prove n is odd.

Here is the proof.

1

Assume that n is even.

2

So n = 2k for some integer k.

3

So n2 = 4k 2 .

4

So n2 is even.

5

This contradicts our initial assumption that n2 is odd.

6

So our assumption that n is even must be false.

7

So n is odd.

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Multiple Strategies

There are often multiple ways to prove something. For instance, consider the following exercise.

Assume n + m is odd. Prove that n × m is even.

We’ll first prove this using indirect proof, then proof by cases.

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1

Assume n × m is odd.

2

Assume n is even.

3

Then n × m is even, since the product of any even number with any number is even.

4

This contradicts our assumption at line 2 that n is even, so n is odd.

5

Assume m is even.

6

Then n × m is even, since the product of any number with any even number is even.

7

This contradicts our assumption at line 5 that m is even, so m is odd.

8

The sum of any two odd numbers is even.

9

So n + m is even.

10 11 12

This contradicts our assumption that n + m is odd.

So our assumption that n × m is odd must be false.

So n × m is even.

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We begin with four mathematical facts.

If n is odd and m is odd, then n + m is even and n × m is odd.

If n is odd and m is even, then n + m is odd and n × m is even.

If n is even and m is odd, then n + m is odd and n × m is even.

If n is even and m is even, then n + m is even and n × m is even.

Now the proof.

1

If n + m is odd, then from the above list of facts we see that either n is odd and m is even, or n is even and m is odd.

2

Assume first that n is odd and m is even.

3

Then, from the second of the above facts, n × m is even.

4

Alternatively, assume that n is even and m is odd.

5

Then, from the third of the above facts, n × m is even.

6

Either way, n × m is even.

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For Next Time

Do the homework: 4.20-4.23, 4.26-4.30 Read Chapter 6

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