PHIL 220. Introduction to Logic
We assign truth values to the propositional variables, and we will use truth functions to interpret the connectives.
\[
\begin{array}{|c|c|}
\hline
\varphi & \neg \varphi \\
\hline
T & F \\
F & T \\
\hline
\end{array}
\]
\(\neg \varphi\) is true iff \(\varphi\) is false.
\((\varphi \wedge \psi)\) is true iff
\(\varphi\) and \(\psi\) are true.
\((\varphi \vee \psi)\) is true iff
\(\varphi\) or \(\psi\) are true.
In English, we may sometimes be inclined to judge a disjunction false if both disjuncts are true, e.g., consider an utterance of the sentence ‘You can have ice cream or you can have cake’.
We may sometimes judge the statement to be false if we can indeed have both ice cream and cake.
We discount the exclusive interpretation of the English connective ‘or’ as a pragmatic effect.
\((\varphi \vee \psi)\) is never false when both \(\varphi\) and \(\psi\) are true.
\((\varphi \to \psi)\) is true iff
\(\varphi\) is false or \(\psi\) is true.
‘If today is a Wednesday, then tomorrow is a workday’
Utterances of the sentence are true throughout the week.
On a Wednesday, both antecedent and consequent are true.
On a Thursday, the antecedent is false but the consequent is true.
On a Friday, both antecedent and consequent are false.
In English, we are often inclined to judge a conditional false when the antecedent is false, e.g., consider an utterance of the sentence: ‘If I scratch my head now, then I will win the lottery tomorrow’.
We may be inclined judge the statement to be false because there is no link (causal or otherwise) between antecedent and consequent.
We discount the apparent requirement for a link between the antecedent and the consequent of conditional in English as a pragmatic effect.
\((\varphi \to \psi)\) is never false when \(\varphi\) is false.
Because truth tables become more complex as more propositional letters are involved in the formula, we need a systematic method for constructing them.
List, in alphabetical order, each distinct propositional letter occurring in the formula. Draw one column for each letter.
Given a complex formula like \((p \to ((q \vee \neg r) \to t)\), we begin:
\[ \begin{array}{|c|c|c|c|} \hline p & q & r & (p \to ((q \vee \neg r) \to q)) \\ \hline & & & & \\ \end{array} \]
Since we have \(3\) propositional letters in our example, we build \(2^3\), namely \(8\) rows.
\[
\begin{array}{|c|c|c|c|}
\hline
p & q & r & (p \to ((q \vee \neg r) \to q)) \\
\hline
& & & \\
\hline
& & & \\
\hline
& & & \\
\hline
& & & \\
\hline
& & & \\
\hline
& & & \\
\hline
& & & \\
\hline
& & & \\
\hline
\end{array}
\]
We now fill the letter-columns systematically:
\[ \begin{array}{|c|c|c|c|} \hline p & q & r & (p \to ((q \vee \neg r) \to q)) \\ \hline T & T & T & \\ \hline T & T & \color{red}{F} & \\ \hline T & \color{red}{F} & T & \\ \hline T & \color{red}{F} & \color{red}{F} & \\ \hline \color{red}{F} & T & T & \\ \hline \color{red}{F} & T & \color{red}{F} & \\ \hline \color{red}{F} & \color{red}{F} & T & \\ \hline \color{red}{F} & \color{red}{F} & \color{red}{F} & \\ \hline \end{array} \]
Draw additional columns for the relevant subformulas, working from the smallest subformulas outward to the complete formula.
Fill each additional column according to the truth-value rule for its main connective.
A formula is a tautology iff it is true under every assignment.
We check if the formula comes out true in every row of the truth table.
The formula
\[ (p \to (p \to q)) \] is not a tautology because it is false in the second row of the truth table.
We check if the formula comes out true in all rows of the truth table.
The formula:
\[ ((p \to q) \vee (q \to p))) \]
is a tautology because it is true in all rows of the truth table.
Two formulas are equivalent iff they are true under exactly the same assignments.
We check if the two formulas agree in truth value in all rows of the truth table.
The two formulas:
\[ \begin{array}{ccc} (p \to q) & & (\neg q \to \neg p) & & \end{array} \]
are equivalent because they agree in truth value in all rows of the table.
We check if the two formulas agree in truth value in all rows of the truth table.
The two formulas:
\[ \begin{array}{ccc} (p \to q) & & (\neg p \to \neg q) & & \end{array} \]
are not equivalent because they differ in truth value in the second and third row of the table.
A set of formulas is consistent iff some assignment makes true every member of the set.
We check if every formula comes out true in some row of the truth table.
The set of three formulas:
\[ \{ (p \to q), (\neg q \to \neg p), \neg p \} \]
is consistent because all three are true in the last two rows of the truth table.
We check if every formula comes out true in some row of the truth table.
The set of four formulas:
\[ \{ (p \to q), (\neg p \to \neg q), \neg q, p \} \]
is inconsistent because no row verifies all four formulas.
An argument is valid iff there is no assignment on which the premises come out true and the conclusion false.
Or, equivalently:
An argument is valid iff every assignment on which the premises are true is one on which the conclusion is true.
We check that no row of the truth table makes the premises true and the conclusion false.
The argument:
\((p \to q)\)
\((q \to \neg r)\)
\(r\)
\(\neg p\)
is valid because no row makes the premises true and the conclusion false.
The argument:
\((p \to q)\)
\((\neg p \to \neg q)\)
\((q \vee r)\)
\(r\)
is invalid because the second row makes the premises true and the conclusion false.