natural deduction: conjunction

PHIL 220. Introduction to Logic

Gabriel Uzquiano

natural deduction: conjunction

Truth tables provide a decision method for propositional validity. They enable us to decide in a finite number of steps whether a given argument in propositional logic is valid:

  • If the argument is valid, there is no row of the truth table on which the premises come out true and the conclusion false.

  • If the argument is invalid, there is at least one row in which the premises come out true and the conclusion is false.

    There is a counterexample to the validity of the argument.

the limits of truth tables

Unfortunately, truth tables are not helpful when it comes to more complex formal languages for which we will have resort to a different strategy.

The other strategy for propositional validity involves proofs:

If an argument is valid, then there is a proof of the conclusion from the premises, which consists of a chain of simple inferences governed by sanctioned rules of inference associated with each connective.

the plan

  1. Interest rates are high.
  2. If interest rates are high and economic growth is slow, then inflation will come down.
  3. Economic growth is slow.
  1. Inflation will come down.

We break down the argument into two simple steps:

  • from 1 and 3:

    interest rates are high and economic growth is slow.

  • from 2 and the above:

    inflation will come down.

  • One step proceeds from premises 1 and 3 and is sanctioned by the semantic behavior of and:

    given \(\varphi\) and \(\psi\), we may infer the conjunction \(\varphi\) and \(\psi\).

  • The other step proceeds from premise 2 and the outcome of the prior step and is sanctioned by the behavior of if … then: given a conditional

    if \(\varphi\), then \(\psi\)

    and

    \(\varphi\),

    we are free to conclude \(\psi\).

translation into propositional logic

\(p\): Interest rates are high.

\(q\): Economic growth is slow.

\(r\): Inflation will come down.

  1. \(p\)
  2. \((p \wedge q) \to r\)
  3. \(q\)
  1. \(r\)
  • Lines 1, 2, and 3 are premises

  • Line 4 is obtained from lines 1 and 3 by \(I\wedge\):

    from \(\varphi\) and \(\psi\), infer \(\varphi \wedge \psi\)

  • Line 5 is obtained from lines 2 and 4 by the rule of inference \(E\to\):

    from if \(\varphi\), then \(\psi\) and \(\varphi\) infer \(\psi\)

  • Line 5 completes the proof because the formula is exactly the conclusion we are after.

what is a proof?

A proof in propositional logic is a finite sequence of formulas each of which is either

  • an assumption, or

  • the outcome of an application of a sanctioned rule of inference to prior formulas.

The last formula is the conclusion.

If an argument of propositional logic is valid, then there is a proof of its conclusion from its premises.

a catalogue of rules of inference

Each connective comes with at least two rules of inference:

  • an introduction rule:

    when you may write a formula with that connective as main connective

  • an elimination rule:

    when you may exploit one that is already available from a prior line

Before we look at them, there is one structural rule to our disposal.

repetition

There is just one structural rule.

repetition

You may repeat a formula \(\varphi\) on a line if it is available on a prior line of the derivation.

The simplest proof you will ever write:

  1. \(p\)
  1. \(p\)

\[ p \vdash p \]

conjunction

Two rules:

  • one introduction rule explains when you may write a conjunction \(\varphi \wedge \psi\) on the basis of prior lines.

  • one elimination rule explains how to exploit a conjunction \(\varphi \wedge \psi\) from a prior line.

introduction rule: \(I\wedge\)

  1. \(\varphi\)
  2. \(\psi\)
  1. \(\varphi \wedge \psi\)

You may write \(\varphi \wedge \psi\) if \(\varphi\) and \(\psi\) are both available at prior lines.

how to apply \(I \wedge\)

  1. \(p\)
  2. \(q\)
  1. \(p \wedge q\)

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

one more example

  1. \(p\)
  2. \(q\)
  3. \(r\)
  1. \((p \wedge r) \wedge q\)

\[ p, q, r \vdash (p \wedge r) \wedge q \]

elimination rule: \(E\wedge\)

  1. \(\varphi \wedge \psi\)
  1. \(\varphi\)
  1. \(\varphi \wedge \psi\)
  1. \(\psi\)

You may write either conjunct of a conjunction available at a prior line.

how to apply \(E \wedge\)

  1. \(p\)
  2. \(q \wedge r\)
  1. \(p \wedge r\)

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

both rules in action

  1. \((p \wedge q) \wedge (r \wedge t)\)
  1. \((p \wedge r) \wedge (q \wedge t)\)

\[ (p\wedge q) \wedge (r \wedge t) \vdash (p \wedge r) \wedge (q \wedge t) \]

practice

Three derivations, given only their assumptions.

exercise 1

exercise 2

exercise 3

next time

We will now provide introduction and elimination rules for the conditional and disjunction.