# Philosophy 112

# What is This?

This is a supplement to sections 1.6 and 1.7 of the Logic Text. You should read those *before* you read this.

# Three Kinds of Derivation

Our system of derivation includes three different kinds of derivation—that is, three different ways of boxing and canceling.

**Direct Derivation** (**DD**): box and cancel by deriving the sentence that is on the show line.

Show

⋮

**Conditional Derivation** (**CD**): box and cancel by *deriving the consequent* of the conditional on the show line. You are also allowed to *assume the antecedent* of the conditional on the show line.

Show \({\mathbin{\rightarrow}}\)

ass cd

⋮

**Indirect Derivation** (**ID**): box and cancel by *deriving a contradiction*. You are also allowed to *assume the negation* of the sentence on the show line.^{1}

Show

\({\mathord{\sim}}\) ass id

⋮

\({\mathord{\sim}}\)

Each of these forms of derivation models a natural form of reasoning.

# Conditional Derivation

Here is an example of the sort of reasoning we can model using **Conditional Derivation**:

- If the wolves are hunted to extinction, the deer population will explode.
- If the deer population explodes, the vegetation will be overgrazed.
- If the vegetation is overgrazed, forest fires will become more common.
- Therefore, if the wolves are hunted to extinction, forest fires will become more common.

Here is a scheme of abbreviation:

\(W\) | The wolves are hunted to extinction | \(V\) | The vegetation is overgrazed |

\(P\) | The deer population explodes | \(R\) | Forest fires become common |

Before reading on, try, on a piece of paper, to symbolize each sentence of the argument, using this scheme.

::: {.answers} Here is the argument, symbolized:

| \(\ \ W\mathbin{\rightarrow}P\) | \(\ \ P\mathbin{\rightarrow}V\) | \(\ \ V\mathbin{\rightarrow}R\) | \(\therefore\ W\mathbin{\rightarrow}R\) :::

If we try to construct a direct derivation, we meet a dead end:

1.Show\(W{\mathbin{\rightarrow}}R\)

2.\(W{\mathbin{\rightarrow}}P\)pr

3.\(P{\mathbin{\rightarrow}}V\)pr

4.\(V{\mathbin{\rightarrow}}R\)pr

What moves can we make on line (5)? We can apply DNI to one of the premises, but that won’t get us anywhere. Neither will R. We don’t have what we need to apply MP or MT.

How might we reason from the premises to the conclusion, informally?

Assume for the sake of argument, that the wolvesFrom that assumption, and our first premise, it follows, by MP, that the deer population will explode. And from that, and our second premise, it follows that the vegetation will be overgrazed. And from that, and our third premise, it follows that forest fires will become common. So,arehunted to extinction.on the assumption that* wolves are hunted to extinction, forest fires become common.

A bit more formally,

::: {.boxed} **Show**: If the wolves are hunted to extinction, then forest fires will become common.

**Proof**:

- The wolves are hunted to extinction (assume for the sake of argument)
- If the deer population explodes, the vegetation will be overgrazed (premise)
- So, the vegetation will be overgrazed (from (1) and (2) by Modus Ponens)
- If the vegetation is overgrazed, forest fires will become more common (premise)
- So, forest fires will become more common (from (3) and (4) by Modus Ponens)

This shows that if (1) is granted, then (5) follows, and so establishes the truth of the conditional. :::

We can represent this exact form of reasoning as a derivation. The first step, after entering the show line, is to make the assumption, for the sake of argument, that the antecedent true:

1.Show\(W{\mathbin{\rightarrow}}R\)

Here, ‘ASS CD’ is short for ‘Assume, for the sake of a Conditional Derivation’. We can then use that assumption, together with our premises, to derive the *consequent*, \(R\):

1.Show\(W{\mathbin{\rightarrow}}R\)

2.\(W\)ASS CD

3.\(W{\mathbin{\rightarrow}}P\)PR

4.\(P\)2,3 MP

5.\(P{\mathbin{\rightarrow}}V\)PR

6.\(V\)5,6 MP

7.\(V{\mathbin{\rightarrow}}R\)PR

This derivation demonstrates that *if* \(W\) is assumed, *then* \(R\) follows. And that is what we are trying to show: \(W\mathbin{\rightarrow}R\). So we box and cancel by **Conditional Derivation** (**CD**), pointing to line 8, where we successfully derived the consequent:

1.Show\(W{\mathbin{\rightarrow}}R\)

2.\(W\)ASS CD

3.\(W{\mathbin{\rightarrow}}P\)PR

4.\(P\)2,3 MP

5.\(P{\mathbin{\rightarrow}}V\)PR

6.\(V\)5,6 MP

7.\(V{\mathbin{\rightarrow}}R\)PR

8.\(R\)6,7 MP

So that is what a **Conditional Derivation** looks like. Go back up to the top of this page, and the schematic presentation of CD. Can you see how this derivation fits that pattern?

::: {.answers} Try plugging \(W\) into the box and \(R\) into the circle. :::

Please, if you have not already, go do the exercises in the Logic Text, at the end of section 1.6. The answers are provided at the end of the chapter, so you can check your work. If you get wrong answer and don’t understand why it is wrong, reread section 1.6 and this supplement with an eye to figuring out what you missed.

The following derivation is wrong. Why?

1.Show\(P{\mathbin{\rightarrow}}R\)

2.\(R\)ASS CD

3.\(R{\mathbin{\rightarrow}}P\)PR

4.\(P\)2,3 MP

::: {.answers} The sentence on line (2) is not the antecedent of the conditional on the Show line, and the sentence on line (4) is not the consequent. To derive the conditional by CD, you need to show that, if we assume the *antecedent*, we can derive the *consequent*. You can’t do it the other way around. :::

Translate the bad derivation just given into English, using the following scheme of abbreviation:

\(R\) | Romney is President |

\(P\) | A Republican is President |

Why is this form of reasoning invalid?

::: {.answers} Here is the derivation, in English:

1.ShowIf a Republican is President, then Romney is President

2.Romney is Presidentass cd

3.If Romney is President, then a Republican is Presidentpr

4.A Republican is President2 3 mp

If we were allowed to do this, we could infer from the premise,

- If Romney is President, then a Republican is.

to the conclusion,

- If a Republican is President, then Romney is.

But that is not a valid inference! Some other Republican could be President. :::

Conditional Derivation is a powerful tool! You should use it when you can. You can use it whenever the show line is a conditional.

::: {.boxed} **Strategic Advice**: If the **show line** is a conditional, try to construct a **conditional derivation**. :::

# Indirect Derivation

Section 1.7 introduces **Indirect Derivation** (**ID**). **ID** models a powerful form of reasoning, common in mathematics and philosophy, that is often called *reductio ad absurdum* (Latin for “reduce to absurdity”).

Let’s begin by considering two real examples of arguments by reductio, one from philosophy and one from mathematics, and one toy example.

## First Example: the Euthyphro

Here is an example drawn from Plato’s dialogue, *The Euthyphro*:

Euthyphro: What is loved by the gods is pious, and what is not loved by the gods is impious.

[…]

Socrates: Haven’t we also said that the gods quarrel and differ with one another, and that there’s mutual hostility among them?

Euthyphro: Indeed, we did say that.

[…]

Socrates: Then the same things, it seems, are both hated and loved by the gods […]

Euthyphro: It seems that way.

Socrates: So, on your account, Euthyphro, the same things would be both pious and impious.

Euthyphro: Apparently.

What is going on here? Euthyphro proposes a definition or account of piety:

\(P\): What is loved by the gods is pious, and what is not loved by the gods is impious.

Socrates wants to show that this is wrong. So,

::: {.boxed} **Show**: \(\mathord{\sim}P\): It is not the case that what is loved by the gods is pious and what is not loved by the gods is impious.

**Proof**:

- Suppose, for reductio, that \(P\) is true.
- The gods often disagree and quarrel (premise).
- Gods disagree and quarrel only if the same things are both hated by some gods and loved by other gods (premise).
- So the same things are both hated by some gods and loved by other gods (from (2) and (3))
- So the same things are both pious and impious (from (1) and (4)).

But (5) is a contradiction, and so false: the same thing cannot both be pious and not pious.

Assuming our reasoning from (1), (2) and (3) to (5) is valid, and given that (5) is false, it follows that at least one of (1), (2), or (3) is false. Since (2) and (3) are granted as premises, it follows that (1) is false, and so that \(\mathord{\sim}P\) is true. :::

This is a complicated argument. I urge you to go read the *Euthyphro* yourself, and see whether or not you think that Socrates’s reasoning is valid. But here we are just interested in the structure of the reasoning. Can you make it fit the scheme for ID?

Show\({\mathord{\sim}}\)

ass id

⋮

\({\mathord{\sim}}\)

As a first step, plug Euthphryo’s proposal, \(P\), into . What should we plug into ? This is tricky. We need two sentences, one the negation of the other, that together capture line (5) of the argument. We can approximate this if we suppose that \(x\) is something that some gods love and some gods hate, and let \(T\) stand for ‘\(x\) is pious’.

## Second Example: Euclid’s Theorem

Here is another famous argument by reductio: Euclid’s proof that there are infinitely many prime numbers.

::: {.boxed} **Show**: There are infinitely many prime numbers

**Proof**:

- Suppose, for reductio, that there are only finitely many primes.
- If there are only finitely many primes, then for some finite \(n\), we list all the primes as \(p_1\) through \(p_n\).
- \(p_1\) through \(p_n\) are all the primes.
- Let \(P = (p_1 \times p_2 \times ... p_n) + 1\)
- \(P\) is not a multiple of any of \(p_1\) … \(p_n\): for any \(k\) between 1 and \(n\), \(\frac{P}{p_k}\) leaves a remainder of 1.
- So the prime factors of P are not on the list.
- So \(p_1\) through \(p_n\) are not all the primes.

But (7) contradicts (3). So the assumption we made for reductio must be false: there are not finitely many primes. :::

A rigorous presentation of Euclid’s proof would require a careful defense of (5). But here I just want you to think about the structure of the proof: to show that there are infinitely many prime numbers, Euclid argues that the opposite assumption leads to a contradiction. Can you fit this into our schema for ID?

## Third Example: Hawk and Dove

Here is a toy example that lends itself to easy representation in our logic.

Suppose Hawk says,

If we invaded Iraq, then we brought stability to the Middle East.

Dove replies with the following argument:

We invaded Iraq. We did not bring stability to the Middle East. \(\therefore\ \) It is not the case that if we invaded Iraq, then we brought stability to the Middle East.

Hawk, let’s suppose, asks Dove to explain why (3) follows from (1) and (2). Dove says:

Assume you are right, and it is true that if we invaded, then we brought stability. Given that we invaded, it follows from your assumption that we brought stability. But we didn’t bring stability. So your assumption leads to a contradiction, and must be rejected.

How do we represent Dove’s line of reasoning?

::: {.boxed} **Show**: It is not the case that if we invaded Iraq, then we brought stability to the Middle East.

**Proof**:

Suppose otherwise:

- If we invaded Iraq, then we brought stability to the Middle East.
- We invaded Iraq (premise).
- We did not bring stability to the Middle East (premise).
- We brought stability to the Middle East (1 2 by Modus Ponens)

(3) and (4) are a contradiction: one of them must be false. So either one of our premises must be false or Hawk’s claim must be false. So, assuming the premises are true, Hawk’s claim is false. :::

We can represent this same line of reasoning as a derivation. First, we can symbolize Dove’s argument:

Now, we set up the derivation,

1.Show\({\mathord{\sim}}(Q{\mathbin{\rightarrow}}S)\)

2.\(Q{\mathbin{\rightarrow}}S\)ass id

3.\(Q\)pr

4.\({\mathord{\sim}}S\)pr

On line (1), Dove states the conclusion to be shown. On line (2), Dove assumes the opposite—the justification is ‘ass id’ which stands for ‘assume for indirect derivation’. On lines (3) and (4), Dove brings down the premises. On line (5), Dove derives \(S\) from (2) and (3).

Lines (4) and (5) are a contradiction. So Dove can box and cancel by **indirect derivation** (**id**), referencing the two lines that contradict each other:

1.Show\({\mathord{\sim}}(Q{\mathbin{\rightarrow}}S)\)

2.\(Q{\mathbin{\rightarrow}}S\)ass id

3.\(Q\)pr

4.\({\mathord{\sim}}S\)pr

5.\(S\)2 3 mp

## Contradictions

Note that ‘contradiction’ is a technical term, and it has a precise syntactic meaning:

::: {.boxed} A **contradiction** is two lines, one the negation of the other.

- If is a sentence, then the pair of sentences, and \(\mathord{\sim}\), are a contradiction. :::

So, using this definition, which of these sentence pairs, when symbolized, are contradictions and which are not?

- It is raining. It is not raining.
- Some gods hate me. Some gods love me.
- I am pious. I am not pious.
- I am happy. I am unhappy.
- The ball is red. The ball is purple.

::: {.answers} a. This is a contradiction. We can symbolize the two sentences as \(R\) and \(\mathord{\sim}R\).

This is not a contradiction: both sentences can be true together. We might symbolize the pair as \(P\) and \(Q\).

This is a contradiction. We might symbolize it as \(P\) and \(\mathord{\sim}P\).

This is one is unclear. Can the same person be both happy and unhappy at the same time? If so, this is not a contradiction. But if being unhappy just means “not being happy”, then this is a contradiction.

This is not a contradiction, even though both sentences cannot be true at the same time (assuming we are talking about the same ball in both sentences). :::

Again, which of these pairs of symbolic sentences are contradictions and which are not?

- \(P\) . \(\mathord{\sim}P\)
- \(P\) . \(\mathord{\sim}\mathord{\sim}P\)
- \(P\) . \(\mathord{\sim}\mathord{\sim}\mathord{\sim}P\)
- \(\mathord{\sim}P\mathbin{\rightarrow}Q\) . \(P\mathbin{\rightarrow}Q\)

::: {.answers} a. Yes, this is a contradiction. b. No, this is not a contradiction. Try fitting it into the pattern we used to define contradictions.

c. No, this is not a contradiction. Try fitting it into the pattern. If you put \(P\) in the box, its contradiction is \(\mathord{\sim}P\), not \(\mathord{\sim}\mathord{\sim}\mathord{\sim}P\). d. This is not a contradiction. The first sentence is in informal notation. In official notation, it looks like this: \((\mathord{\sim}P\mathbin{\rightarrow}Q)\), which makes it clear that it is not the negation of the second sentence. :::

# What Next?

Go the Logic Software, and complete the homework assignment.

If the show line is a negation, you can assume its unnegation instead.↩