natural deduction: negation

PHIL 220. Introduction to Logic

Gabriel Uzquiano

natural deduction: negation

We have introduction and elimination rules for conjunction, conditional, and disjunction. We now introduce rules for negation.

a marker for contradiction: \(\bot\)

To reason with negation in the course of a proof, we introduce a marker we use to record the fact that we have arrived at a contradiction. Think of \(\bot\) not as part of the vocabulary of propositional logic, but rather as a punctuation device we use for purposes of proofs.

We will introduce four rules for negation:

• negation introduction I\(\neg\)
• negation elimination E\(\neg\)
• ex falso sequitor quodlibet EFSQ
• double negation \(\neg \neg\)

elimination rule: E\(\neg\)

1. \(\varphi\)
2. \(\neg \varphi\)

3. \(\bot\)

You may write \(\bot\) if both a formula \(\varphi\) and its negation \(\neg \varphi\) are available from prior lines.

The rule explains that we may write use the marker \(\bot\) to record the fact that we have arrived at a contradiction.

introduction rule: I\(\neg\)

You may write \(\neg \varphi\) if you can prove \(\bot\) from the assumption that \(\varphi\).

Once you complete the subproof, bracket the lines of that subproof and never appeal to them again.

rationale for I\(\neg\)

1. If my client did it, then she walked into the library at 9AM.
2. If my client walked into the library at 9AM, then she was not seen at the garden at 9AM.
3. My client was seen at the garden at 9AM.

3. My client did not do it.

How should we argue for that conclusion?

1. Assume that my client did it.
2. Reason within that assumption:
   • my client walked into the library at 9AM (from 1 and assumption)
   • my client was not seen at the garden at 9AM (from 2 and prior line)
   • my client was seen at the garden at 9AM (from 3)
   • \(\bot\) (from E\(\neg\) and prior lines)
3. Exit the scope of the assumption to conclude:
   • it is not the case that my client did it.

how to combine E\(\neg\) and I\(\neg\)

1. \(p \to q\)
2. \(\neg q\)

2. \(\neg p\)

\[ p \to q, \neg q \vdash \neg p \]

• line 3 opens the subproof
• line 6 closes it and discharges the assumption.

two more examples

\[ p \to \neg p \vdash \neg p \]

\[ \neg (p \vee q) \vdash \neg p \wedge \neg q \]

ex falso sequitor quodlibet: EFSQ

Negation is a special connective in that its introduction and elimination rules are not sufficient to account for all valid arguments in which the connective occurs.

One further rule is EFSQ:

1. \(\bot\)

2. \(\varphi\)

You may write any formula once an occurrence of \(\bot\) becomes available at a prior line.

The Latin phrase ex falso sequitur quodlibet just means that anything at all follows from a contradiction, and we will use the initials EFSQ to abbreviate the rule.

how to apply EFSQ

1. \(p\)
2. \(\neg p\)

3. \(q\)

This argument is valid: no assignment can make the premises true and the conclusion false (simply because no assignment can make the premises true.)

So, we should be able to prove the conclusion from its premises …

\[ p, \neg p \vdash \neg q \]

two more examples

\[ p \vee q, \neg p \vdash q \]

\[ p \wedge q, \neg p \vdash r \]

double negation: \(\neg \neg\)

1. \(\neg \neg \varphi\)

2. \(\varphi\)

You may write a formula \(\varphi\) once its double negation \(\neg \neg \varphi\) becomes available at a prior line.

how to apply \(\neg \neg\)

1. \(\neg p \to \neg q\)
2. \(q\)

3. \(p\)

\[ \neg p \to \neg q, q \vdash p \]

practice

worked example

All four rules, on an argument from natural language.

a natural language argument

If you flip the coin, then it will land heads or tails but not both. I’ll win only if the coin lands heads. However, you’ll win only if the coin lands tails. As a matter of fact, you’ll flip the coin. So, we will not both win.

in premise-conclusion form

1. If you flip the coin, then it will land heads or tails but not both
2. I will win only if the coin lands heads.
3. You will win only if the coin lands tails
4. You will flip the coin.

5. We will not both win.

translation

\(p\): You flip the coin.

\(q\): The coin lands heads.

\(r\): The coin lands tails.

\(s\): I will win.

\(t\): You will win.

1. \(p \to ((q \vee r) \wedge \neg (q \wedge r))\)
2. \(s \to q\)
3. \(t \to r\)
4. \(p\)

4. \(\neg (s \wedge t)\)

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The argument is valid because there is a proof of the conclusion from its premises.