validity

PHIL 220. Introduction to Logic

Gabriel Uzquiano

validity

We aim for a framework within which to assess the validity of argument forms exemplified by natural language arguments.

from English to propositional logic

We will extract a propositional form from each natural language argument:

argument

  1. If mica is a metal, then it conducts electricity.
  2. Mica does not conduct electricity.
  1. Mica is not a metal.

argument form

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

key

\(p\) : Mica is a metal

\(q\) : Mica conducts electricity

Is this a valid argument form?

validity

An argument of propositional logic is valid iff there is no assignment of truth values to its propositional variables on which the premises are true and the conclusion is false.

from English to propositional logic

argument

  1. If the government levies another tax on companies, then employment will fall.
  2. The government will not levy another tax.
  1. Employment will not fall.

argument form

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

key

\(p\) : The government levies another tax on companies.

\(q\) : Employment falls.

Is this a valid argument form?

validity

counterexamples to validity

Given an invalid argument, a counterexample to validity is an assignment of truth values to propositional variables on which the premises come out true and the conclusion comes out false.

invalid argument form

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

invalid argument form

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

counterexample

\(p\) : \(F\)

\(q\) : \(T\)

counterexample

\(p\) : \(F\)

\(q\) : \(T\)

\(r\) : \(T\)

how to search for a counterexample

One way to check the validity of an argument is to scan a full truth table, row by row, in search of a counterexample, but another is to use a partial truth table to search for a counterexample.

This method is particularly helpful for complex arguments that involve more than two propositional variables. We would rather not scan 16 or 32 rows of a full truth table for an argument in which 4 or 5 propositional variables are involved.

the search for a counterexample method

Consider the argument form:

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

To search for a counterexample, follow the steps below:

  1. Focus on assignments that make the conclusion false

    For \(\neg r \to q\) to be false, \(\neg r\) must be true and \(q\) must be false. Therefore, we look at assignments on which both \(r\) is false and \(q\) is false.

  2. Ask which of the assignments that remain make the premises true.

    There are just two options for \(r\) is false and \(q\) is false, namely, assignments on which \(p\) is true and assignments on which \(p\) is false. If \(p\) is true and \(r\) and \(q\) are false, then premise 1 will be true because \(q \to r\) comes out true and premise 2 will be true because \(\neg r \to p\) has a true consequent.

  3. If you find a counterexample, then the argument is invalid.

    Make \(p : T\), \(q : F\), \(r : F\). The assignment \(T F F\) is a counterexample to validity.

example 1

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

There are \(2^3 = 8\) assignments.

A full truth table would require 8 rows.

We now employ the search for counterexample method:

  1. Find assignments that make the conclusion false

To make

\[ q \to \neg r \]

false, we must have:

\[ q : T \qquad r : T \]

So, we need consider only 2 rows out the 8:

\[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & r & p \to ((q \wedge \neg r) \vee (r \wedge q)) & p \wedge q & r \vee \neg p & q \to \neg r \\ \hline T & T & T & & & & \color{red}{F} \\ F & T & T & & & & \color{red}{F} \\ \hline \end{array} \]

filter the candidate rows

  1. Ask which of the assignments that remain make the premises true.

To make the second premise true, we focus on the first row:

\[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & r & p \to ((q \wedge \neg r) \vee (r \wedge q)) & p \wedge q & r \vee \neg p & q \to \neg r \\ \hline T & T & T & &\color{red}{T} & & \color{red}{F} \\ F & T & T & & F & & \color{red}{F} \\ \hline \end{array} \]

Check the other two premises now:

\[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & r & p \to ((q \wedge \neg r) \vee (r \wedge q)) & p \wedge q & r \vee \neg p & q \to \neg r \\ \hline T & T & T & \color{red}{T} &\color{red}{T} & \color{red}{T} & \color{red}{F} \\ F & T & T & & F & & \color{red}{F} \\ \hline \end{array} \]

We conclude that \(T T T\) is a counterexample.

The argument is invalid.

truth table

Consider the assignment \(p : T,\ q : T,\ r : T\).

example 2

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

There are \(2^3 = 8\) assignments.

A full truth table would require 8 rows.

We now employ the search for a counterexample method:

  1. Find assignments that make the conclusion false

To make

\[ p \to r \]

false, we must have:

\[ p : T \qquad r : F \]

So, we need consider only 2 rows out the 8:

\[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & r & p \vee q & \neg r \to q & p \to r \\ \hline T & T & F & & & \color{red}{F} \\ T & F & F & & & \color{red}{F} \\ \hline \end{array} \]

filter the candidate rows

  1. Ask which of the assignments that remain make the premises true.

Both rows verify the first premise:

\[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & r & p \vee q & \neg r \to q & p \to r \\ \hline T & T & F & \color{red}{T} & & \color{red}{F} \\ T & F & F & \color{red}{T} & & \color{red}{F} \\ \hline \end{array} \]

Check the other premise now:

\[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & r & p \vee q & \neg r \to q & p \to r \\ \hline T & T & F & \color{red}{T} & \color{red}{T} & \color{red}{F} \\ T & F & F & \color{red}{T} & F & \color{red}{F} \\ \hline \end{array} \]

We conclude that \(T T F\) is a counterexample.

The argument is invalid.

truth table

Consider the assignment \(p : T,\ q : T,\ r : F\).

from English to propositional logic

argument

  1. If the government levies another tax, then employment will fall or inflation will rise.
  2. If inflation rises, then if the government levies another tax, then employment will fall.
  1. Employment will fall or inflation will rise.

argument form

  1. \(p \to (q \vee r)\)
  2. \(r \to (p \to q)\)
  1. \(q \vee r\)

key

\(p\) : The government levies another tax.

\(q\) : Employment falls.

\(r\) : Inflation rises.

in search of a counterexample

  1. Find assignments that make the conclusion false

To make

\[ q \vee r \]

false, we must have:

\[ q : F \qquad r : F \]

So, we need consider only 2 rows out the 8:

\[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & r & p \to (q \vee r) & r \to (p \to q) & q \vee r \ \\ \hline T & T & F & & & \color{red}{F} \\ F & F & F & & & \color{red}{F} \\ \hline \end{array} \]

filter the assignments

  1. Ask which of the assignments that remain make the premises true.

Only the second row verifies the first premise:

\[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & r & p \to (q \vee r) & r \to (p \to q) & q \vee r \ \\ \hline T & F & F & F& & \color{red}{F} \\ F & F & F & \color{red}{T} & & \color{red}{F} \\ \hline \end{array} \]

Check the other premise now:

\[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & r & p \to (q \vee r) & r \to (p \to q) & q \vee r \ \\ \hline T & F & F & F& & \color{red}{F} \\ F & F & F & \color{red}{T} & \color{red}{T} & \color{red}{F} \\ \hline \end{array} \]

We conclude that \(F F F\) is a counterexample to validity. The argument is invalid.

from English to propositional logic

argument

  1. If the coin lands heads, it won’t land tails, and if it lands tails, it won’t land heads.
  2. It’s not the case that I tossed the coin and it landed heads.
  3. It’s not the case that if I tossed the coin, then it landed tails.
  1. I didn’t toss the coin.

argument form

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

key

\(p\) : The coin landed heads.

\(q\) : The coin landed tails.

\(r\) : I tossed the coin.

in search of a counterexample

  1. Find assignments that make the conclusion false

To make

\[ \neg r \]

false, we must have:

\[ r : T \]

So, we need consider only 4 rows out the 8:

\[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & r & (p \to \neg q) \wedge (q \to \neg p) & \neg (r \wedge p) & \neg (r \to q) & \neg r \ \\ \hline T & T & T & & & & \color{red}{F} \\ F & T & T & & & & \color{red}{F} \\ T & F & T & & & & \color{red}{F} \\ F & F & T & & & & \color{red}{F} \\ \hline \end{array} \]

filter the assignments

  1. Ask which of the assignments that remain make the premises true.

Only two of the rows that remain verify the third premise:

\[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & r & (p \to \neg q) \wedge (q \to \neg p) & \neg (r \wedge p) & \neg (r \to q) & \neg r \ \\ \hline T & T & T & & & F & \color{red}{F} \\ F & T & T & & & F & \color{red}{F} \\ T & F & T & & & \color{red}{T} & \color{red}{F} \\ F & F & T & & & \color{red}{T} & \color{red}{F} \\ \hline \end{array} \]

Check the second premise now:

\[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & r & (p \to \neg q) \wedge (q \to \neg p) & \neg (r \wedge p) & \neg (r \to q) & \neg r \ \\ \hline T & T & T & & & F & \color{red}{F} \\ F & T & T & & & F & \color{red}{F} \\ T & F & T & & F & \color{red}{T} & \color{red}{F} \\ F & F & T & & \color{red}{T} & \color{red}{T} & \color{red}{F} \\ \hline \end{array} \]

filter the assignments

Check the first premise now:

\[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & r & (p \to \neg q) \wedge (q \to \neg p) & \neg (r \wedge p) & \neg (r \to q) & \neg r \ \\ \hline T & T & T & & & F & \color{red}{F} \\ F & T & T & & & F & \color{red}{F} \\ T & F & T & & F & \color{red}{T} & \color{red}{F} \\ F & F & T & \color{red}{T} & \color{red}{T} & \color{red}{T} & \color{red}{F} \\ \hline \end{array} \]

We conclude that \(F F T\) is a counterexample to validity. The argument is invalid.