CH3 – Slides

Author

Affiliation

David W. Agler

The Pennsylvania State University

Abstract

Slides for Ch3 of my logic course.

PL: Truth Tables

Slides are available at www.davidagler.com

Informal Test: Example

• P1. Some basketball players are millionaires.
• P2. Some millionaires have fancy cars.
• C. Therefore, some basketball players have fancy cars.

90% will say “valid” but it is invalid.

Informal Test: Example 2

• P1. If my car is out of gas, then it won’t start.
• P2. My care won’t start.
• C. Therefore, it is out of gas.

Many will say “valid” but it is invalid.

Informal tests

Informal tests (logical intuition, logical imagination) have problems

1. Wrong or inconsistent results
2. Belief bias: true conclusion therefore valid argument.
3. Large arguments
4. Abstract content

We want a test that …

1. gives consistent results
2. tests validity without belief bias
3. can be applied to large arguments
4. can be formulated about any subject matter

What is a truth table?

NoteDefinition

A truth table for PL is a table that provides a graphical way of representing valuations of wff(s) under a set of interpretations.

We’ve used truth tables already

\[ \begin{array}{c|c} \phi & \neg (\phi)\\ \hline T & F\\ F & T \end{array} \]

\[ \begin{array}{c c|c c c c} \phi&\psi&\phi\wedge \psi&\phi\vee \psi&\phi \to \psi&\phi\leftrightarrow \psi\\ \hline T&T&T&T&T&T\\ T&F&F&T&F&F\\ F&T&F&T&T&F\\ F&F&F&F&T&T \end{array} \]

Using table to test arguments

1. Gives consistent results
2. Finite mechanical method (no bias)
3. Works on large arguments
4. Works on any subject matter
5. “Is the argument valid?” Always gives “yes” or “no” answer

The plan.

1. Use tables to determine T-value of any wff (under single interpretation)
2. Use tables to determine T-value of any wff (under every interpretation)
3. Use table as a tool to check validity (and other stuff)

Truth Tables: Step by Step

1. Write down the wff.
2. Write the truth value (\(T\) or \(F\)) under each propositional letter in the wff for each interpretation \(\mathscr{I}\).
3. Assign \(T\) or \(F\) to subformulas in the order that the wff is constructed using (1) the values assigned to the subformulas used to construct that wff and (2) the valuation function assigned to the operator added to the subformula constructed.

CautionStep 3?

Step 3 is precise but not intuitive.

Step 1

Let’s create a truth table for this wff: \(P\land \neg Q\)

• Step 1: Write down the wff with each propositional to the left of it

\[ \begin{array}{c c|c c c c} P&Q &P& \land & \neg & Q\\\hline T&F & & & \\ \end{array} \]

Step 2

• Step 2: Consider an interpretation of the propositional letters

\[ \begin{array}{c c|c c c c} P&Q &P& \land & \neg & Q\\\hline T&F & & & \\ \end{array} \]

write the T-values under the letters of \(P\land \neg Q\).

\[ \begin{array}{c c|c c c c} P&Q &P& \land & \neg & Q\\\hline T&F &T & & &F\\ \end{array} \]

Step 3.1: Calculate the truth table

• Step 3.1: Construct the formula using formation rules.

\[ \begin{array}{c c|c c c c} P&Q &P& \land & \neg & Q\\\hline T&F &T & & &F\\ \end{array} \]

1. \(P, Q\) are wffs.
2. If \(Q\) is a wff, then \(\neg Q\) is a wff.
3. If \(P\) and \(\neg Q\) are wffs, then \(P\land \neg Q\) is a wff.

Step 3.2: Calculate the truth table

Assign truth values to the subformulas in the order in which you constructed the formula

1. Assign T-values to \(P\) and \(Q\).
2. Assign T-value to \(\neg Q\)
3. Assign T-value to \(P\land \neg Q\)

Truth Tables for the Operators

\[ \begin{array}{c|c} \phi & \neg (\phi)\\ \hline T & F\\ F & T \end{array} \]

\[ \begin{array}{c c|c c c c} \phi&\psi&\phi\wedge \psi&\phi\vee \psi&\phi \to \psi&\phi\leftrightarrow \psi\\ \hline T&T&T&T&T&T\\ T&F&F&T&F&F\\ F&T&F&T&T&F\\ F&F&F&F&T&T \end{array} \]

Step 3.2: The Negation

\[ \begin{array}{c c|c c c c} P&Q &P& \land & \neg & Q\\\hline T&F &T & & &\color{blue}{F}\\ \end{array} \]

\[ \begin{array}{c|c} \phi & \neg (\phi)\\ \hline T & F\\ F & T \end{array} \]

\[ \begin{array}{c c|c c c c} P&Q &P& \land & \neg & Q\\\hline T&F &T & & \color{blue}{T}&F\\ \end{array} \]

Step 3.2: Illustrated

\[ \begin{array}{c c|c c c c} P&Q &P& \land & \neg & Q\\\hline T&F &\color{blue}{T} & & \color{blue}{T}&F\\ \end{array} \]

\[ \begin{array}{c c|c c c c} \phi&\psi&\phi\wedge \psi&\phi\vee \psi&\phi \to \psi&\phi\leftrightarrow \psi\\ \hline \color{blue}{T}&\color{blue}{T}&\color{blue}{T}&T&T&T\\ T&F&F&T&F&F\\ F&T&F&T&T&F\\ F&F&F&F&T&T \end{array} \]

\[ \begin{array}{c c|c c c c} P&Q &P& \land & \neg & Q\\\hline T&F &T & \color{blue}{T}& T&F\\ \end{array} \]

Reading the Table

What is the truth value of \(P\land \neg Q\)? \[ \begin{array}{c c|c c c c} P&Q &P& \land & \neg & Q\\\hline T&F &T & \color{blue}{T}& T&F\\ \end{array} \]

The truth value of \(P\land \neg Q\) is under its main operator.

Example 2

• “It is NOT the case that Tek is smart iff he’s rich.”
• Translation: \(\neg (S\leftrightarrow R)\)
• Facts: Tek is smart but he’s not rich.
• Interpretation: \(\mathscr{I}(S)=T\), \(\mathscr{I}(R)=F\)

Example 2 - Step 1

Step 1: Write out the formula with the individual letters to the left.

\[ \begin{array}{cc|c c c c} S&R& \neg &(S & \leftrightarrow & R)\\\hline T&F&&&& \end{array} \]

Truth Table: An Example

Next, consider how \(\neg (S \leftrightarrow R)\) is constructed.

1. \(S, R\) are wffs.
2. If \(S, R\) are wffs, then \((S \leftrightarrow R)\) is a wff.
3. If \((S \leftrightarrow R)\) is a wff, then \(\neg (S \leftrightarrow R)\) is a wff.

Assign truth values to the subformulas of \(\neg (S \leftrightarrow R)\) in that order.

Truth Table – An Example

Start by writing truth values under the proposition letters in the wff. Assume \(\mathscr{I}(S)=T\) and \(\mathscr{I}(R)=F\).

\[ \begin{array}{cc|c c c c} S&R& \neg &(S & \leftrightarrow & R)\\\hline T&F&&&&\\ \end{array} \]

Write the T-values under \(S\) and \(R\) under the wff.

\[ \begin{array}{cc|c c c c} S&R& \neg &(S & \leftrightarrow & R)\\\hline T&F&&T&&F\\ \end{array} \]

Example 2

The next wff constructed is \((S \leftrightarrow R)\).

\[ \begin{array}{cc|c c c c} S&R& \neg &(S & \color{blue}{\leftrightarrow} & R)\\\hline T&F&&\color{blue}{T}&&\color{blue}{F}\\ \end{array} \]

To determine whether \((S \leftrightarrow R)\) is T or F, look at the truth table where the leftside of the biconditional is T and the rightside is F:

\[ \begin{array}{c c|c c c c} \phi&\psi&\phi\wedge \psi&\phi\vee \psi&\phi \to \psi&\phi\leftrightarrow \psi\\ \hline T&T&T&T&T&T\\ \color{blue}{T}&\color{blue}{F}&F&T&F&\color{blue}{F}\\ F&T&F&T&T&F\\ F&F&F&F&T&T \end{array} \]

Example 2

\[ \begin{array}{cc|c c c c} S&R& \neg &(S & \color{blue}{\leftrightarrow} & R)\\\hline T&F&&\color{blue}{T}&&\color{blue}{F}\\ \end{array} \]

\[ \begin{array}{cc|c c c c} S&R& \neg &(S & \leftrightarrow & R)\\\hline T&F&&T&\color{blue}{F}&F\\ \end{array} \]

\(v(S \leftrightarrow R)=F\)

Truth Table: An Example

The next wff constructed is \(\neg (S \leftrightarrow R)\).

\[ \begin{array}{cc|c c c c} S&R& \color{blue}{\neg} &(S & \leftrightarrow & R)\\\hline T&F&&T&\color{blue}{T}&F\\ \end{array} \]

• We use \((S \leftrightarrow R)\) to construct \(\neg (S \leftrightarrow R)\)
• So, we will use the T-value of \((S \leftrightarrow R)\) to determine the T-value of \(\lnot (S \leftrightarrow R)\).
• Use the T-value under \(\leftrightarrow\) and the valuation for Negation.

Truth Table: An Example

\[ \begin{array}{cc|c c c c} S&R& \color{blue}{\neg} &(S & \leftrightarrow & R)\\\hline T&F&&T&\color{blue}{F}&F\\ \end{array} \]

\[ \begin{array}{cc|c c c c} S&R& \neg &(S & \leftrightarrow & R)\\\hline T&F&\color{blue}{T}&T&T&F\\ \end{array} \]

\(v\neg(S \leftrightarrow R)=T\)

Example 2

• “It is NOT the case that Tek is smart iff he’s rich.”
• Translation: \(\neg (S\leftrightarrow R)\)
• Facts: Tek is smart but he’s not rich.
• Interpretation: \(\mathscr{I}(S)=T\), \(\mathscr{I}(R)=F\)
• Valuation: \(v\neg(S \leftrightarrow R)=F\)

Let’s Practice 1

• Book, Ex 3.30, p.113 #1

Let’s Practice 2

• Book, Ex 3.30, p.113 #5

Let’s Practice 3

• Book, Ex 3.30, p.113 #10 (this one is challenging)

Next Step

• We can determine if \(\phi\) is T or F under a single interpretation
• We need to determine the truth value of multiple wffs under all admissible interpretations

NoteExam Tip

You’ll need to create a table on the exam! So, practice, practice, practice!

The Truth-Table Method

The Truth-Table Method

• What we know: How to determine T-value of \(\phi\) under a single interpretation
• Next: How to determine the truth value of wffs under all every interpretation

Step 1: Write the wff

Step 1: Write out the wff \(\phi\) and all of the propositional letters in \(\phi\) to the left of \(\phi\).

\[ \begin{array}{c c|c c c c c} P&R&\neg&P&\vee&\neg&R\\ \hline \end{array} \]

Step 2: Every interpretation

Step 2: Write all interpretations for the propositional letters in \(\phi\).

NoteHow many interpretations are there?

The number of possible interpretations (rows) is determined by the number of distinct propositional letters. If \(\phi\) has \(n\) distinct letters, you’ll need \(2^n\) table rows.

• \(P\): \(2^1=2\) rows
• \(P\wedge Q\): \(2^2=4\) rows
• \((P\wedge Q)\wedge R\): \(2^3=8\) rows

1 Propositional letter

Interpretation	\(P\)
\(\mathscr{I}_1\)	\(T\)
\(\mathscr{I}_2\)	\(F\)

2 Propositional letters

Interpretation	\(P\)	\(Q\)	\(P\land Q\)
\(\mathscr{I}_1\)	T	T
\(\mathscr{I}_2\)	T	F
\(\mathscr{I}_3\)	F	T
\(\mathscr{I}_4\)	F	F

3 Propositional letters

Interpretation	\(P\)	\(Q\)	\(R\)	\((P\land Q)\land R\)
\(\mathscr{I}_1\)	T	T	T
\(\mathscr{I}_2\)	T	T	F
\(\mathscr{I}_3\)	T	F	T
\(\mathscr{I}_4\)	T	F	F
\(\mathscr{I}_5\)	F	T	T
\(\mathscr{I}_6\)	F	T	F
\(\mathscr{I}_7\)	F	F	T
\(\mathscr{I}_8\)	F	F	F

4 Propositional letters

Interpretation	\(P\)	\(Q\)	\(R\)	\(S\)
\(\mathscr{I}_1\)	T	T	T	T
\(\mathscr{I}_2\)	T	T	F	T
\(\mathscr{I}_3\)	T	F	T	T
\(\mathscr{I}_4\)	T	F	F	T
\(\mathscr{I}_5\)	F	T	T	T
\(\mathscr{I}_6\)	F	T	F	T
\(\mathscr{I}_7\)	F	F	T	T
\(\mathscr{I}_8\)	F	F	F	T
\(\mathscr{I}_9\)	T	T	T	F
\(\mathscr{I}_{10}\)	T	T	F	F
\(\mathscr{I}_{11}\)	T	F	T	F
\(\mathscr{I}_{12}\)	T	F	F	F
\(\mathscr{I}_{13}\)	F	T	T	F
\(\mathscr{I}_{14}\)	F	T	F	F
\(\mathscr{I}_{15}\)	F	F	T	F
\(\mathscr{I}_{16}\)	F	F	F	F

Tip for covering all interpretations

• Start with the rightmost letter, and alternate: TFTFTFTF
• Double the pattern: TT FF TT FF TT FF
• If necessary, double again: TTTT FFFF TTTT FFFF

Applying Step 2

\[ \begin{array}{c c|c c c c c} P&R&\neg&P&\vee&\neg&R\\ \hline &&&&&&\\ \end{array} \]

\[ \begin{array}{c c|c c c c c} P&R&\neg&P&\vee&\neg&R\\ \hline T&T&&&&&\\ T&F&&&&&\\ F&T&&&&&\\ F&F&&&&& \end{array} \]

Step 3: Transfer the T-values

Step 3: For each row, write the truth values under the corresponding letter.

\[ \begin{array}{c c|c c c c c} P&R&\neg&P&\vee&\neg&R\\ \hline T&T&&T&&&T\\ T&F&&T&&&F\\ F&T&&F&&&T\\ F&F&&F&&&F \end{array} \]

Step 4: Table method for each row

Step 4: Assign \(T\) or \(F\) to subformulas in the order that the wff is constructed.

\[ \begin{array}{c c|c c c c c} P&R&\neg&P&\vee&\neg&R\\ \hline T&T&F&T&&F&T\\ T&F&F&T&&T&F\\ F&T&T&F&&F&T\\ F&F&T&F&&T&F \end{array} \]

Step 4: Continued

\[ \begin{array}{c c|c c c c c} P&R&\neg&P&\vee&\neg&R\\ \hline T&T&F&T&&F&T\\ T&F&F&T&&T&F\\ F&T&T&F&&F&T\\ F&F&T&F&&T&F \end{array} \]

Now determine the truth value of \(\neg P\vee\neg R\).

\[ \begin{array}{c c|c c c c c} P&R&\neg&P&\vee&\neg&R\\ \hline T&T&F&T&F&F&T\\ T&F&F&T&T&T&F\\ F&T&T&F&T&F&T\\ F&F&T&F&T&T&F \end{array} \]

The table shows the truth value of \(\neg P\vee\neg R\) under all interpretations of \(P\) and \(R\) in \(PL\).

Exercise

See book, Ex. 3.31, pp.115, #1 (\(P\to \lnot R\))

Truth-Table Analysis

Truth-Table Analysis

Truth tables can be used to:

1. determine the truth value of a complex wff under a single (or all) interpretations
2. test whether a wff is a contingency, tautology, or contradiction;
3. test whether a pair of wffs are equivalent;
4. test whether a collection of wffs is consistent; and
5. test whether an argument is valid.

Contingency, Tautology, Contradiction

Contradiction

NoteDefinition

A proposition \(\mathbf{P}\) is a contradiction if and only if \(\mathbf{P}\) is always false.

• The number is both even and not even.
• Tek is my friend and not my friend.

NoteDefinition

A wff \(\phi\) is a PL-contradiction if and only if \(v(\phi)=F\) under every interpretation.

Tautology

NoteDefinition

A proposition \(\mathbf{P}\) is a tautology if and only if \(\mathbf{P}\) is always true.

• The student passed or did not pass.
• If it is snowing, then it is snowing.

NoteDefinition

A wff \(\phi\) is a PL-tautology if and only if \(v(\phi)=T\) under every interpretation.

Contingency

NoteDefinition

A proposition \(\mathbf{P}\) is a contingency if and only if its truth value depends upon the nature of the world.

• Tek passed his logic class.
• I am 6’0 tall.

NoteDefinition

A wff \(\phi\) is a PL-contingency if and only if \(\phi\) is neither a contradiction nor a tautology. Equivalently, \(v(\phi)=T\) under at least one interpretation and \(v(\phi)=F\) under at least one interpretation.

Contingency, Etc: Why Care?

• Tautologies are T and contradictions are F in virtue of their form.
• No need to check the real world
• For contingencies, we need to check the real world.

How to Test

Tables can check whether \(\phi\) is a contradiction, tautology, or contingency:

1. Construct the truth table.
2. Check whether the wff is

• F under every interpretation (contradiction),
• T under every interpretation (tautology),
• Neither a contradiction nor a tautology (contingency)

Practice: \(\neg P\vee\neg R\)

Is \(\neg P\vee\neg R\) a PL-contingency, PL-tautology, or PL-contradiction?

\[ \begin{array}{c c|c c c c c} P&R&\neg&P&\vee&\neg&R\\ \hline T&T&F&T&F&F&T\\ T&F&F&T&T&T&F\\ F&T&T&F&T&F&T\\ F&F&T&F&T&T&F \end{array} \]

It is a PL-contingency.

Practice: \(P\vee\neg P\)

Is \(P\vee\neg P\) a PL-contingency, PL-tautology, or PL-contradiction?

\[ \begin{array}{c|c c c c} P&P&\vee&\neg&P\\ \hline T&T&T&F&T\\ F&F&T&T&F \end{array} \]

This wff is a tautology.

Practice: \(P\wedge\neg P\)

Is \(P\wedge\neg P\) a PL-contingency, PL-tautology, or PL-contradiction?

\[ \begin{array}{c|c c c c} P&P&\wedge&\neg&P\\ \hline T&T&F&F&T\\ F&F&F&T&F \end{array} \]

This wff is a contradiction.

Practice: An English Example

1. It is not the case that if Liz does not eat ice cream, then she does not eat cake.
2. Translate as \(\neg(\neg I\rightarrow\neg C)\).
3. Create the table for \(\neg(\neg I\rightarrow\neg C)\).
4. Check whether the wff is a PL-contingency, PL-tautology, or PL-contradiction.

Continued

\[ \begin{array}{c c|c c c c c c c} I&C&\neg&( &\neg&I&\rightarrow&\neg&C& )\\ \hline T&T&F&&F&T&T&F&T&\\ T&F&F&&F&T&T&T&F&\\ F&T&T&&T&F&F&F&T&\\ F&F&F&&T&F&T&T&F& \end{array} \]

• Look under the leftmost negation
• Notice some Ts and some Fs.
• \(\neg(\neg I\rightarrow\neg C)\) is a PL-contingency.
• Therefore, the sentence is a contingency.

Exercise

See book, Ex. 3.33, pp.122, #6: \(\lnot (P\vee \lnot R)\)

Consistency

Consistency

NoteDefinition

A set of propositions is consistent if and only if they all can be true; that is, there is some way that they can all be true.

• We are friends. I like you. It is sunny.
• I will raise taxes. I will improve public services (e.g., better roads, schools, etc.)

NoteDefinition

A set of wffs \(\Gamma\) is PL-consistent if and only if there is at least one interpretation such that all members of \(\Gamma\) are true.

Inconsistency

NoteDefinition

A set of propositions is inconsistent if and only if they cannot all be true at the same time.

• My grade in Logic is good.
• I have a D in Logic.
• To have a good grade in logic, you must have a C or better.

NoteDefinition

A set of wffs \(\Gamma\) is PL-inconsistent if and only if it is not PL-consistent: there is no interpretation such that all members of \(\Gamma\) are true.

Consistency: Why Care?

1. Promises: Can the politician make good on all their promises?
2. Scientific Theory: Newton’s laws of universal gravitation and Mercury’s orbit.
3. Philosophical-Religious Beliefs: God and the existence of evil or God and free will

Consistency: Benefit of a test

In many cases, we can simply see that a set of wffs is PL-consistent.

• \(P\) and \(Q\) are PL-consistent.
• Case: \(\mathscr{I}(P)=T\) and \(\mathscr{I}(Q)=T\).

In other cases, it is not obvious.

• Are \(P\vee Q\) and \(\neg(P\wedge Q)\) PL-consistent?
• What about \(\neg P\vee Q\) and \(\neg(P\wedge Q)\)?
• What about \(\neg P\vee(Q\vee\neg R)\) and \(\neg(P\leftrightarrow Q)\)?

Testing Consistency

1. Write each wff down in a row.
2. Construct a single truth table.
3. Check whether there is at least one row where all the wffs are \(T\).
4. If there is such a row, the wffs are consistent.
5. If there is no such row, the wffs are inconsistent.

Consistency: Example

Test whether \(P\rightarrow Q\), \(P\wedge Q\), and \(P\vee\neg Q\) are PL-consistent.

\[ \begin{array}{c c|c c c|c c c|c c c c} P&Q&P&\rightarrow&Q&P&\wedge&Q&P&\vee&\neg&Q\\ \hline T&T&T&T&T&T&T&T&T&T&F&T\\ T&F&T&F&F&T&F&F&T&T&T&F\\ F&T&F&T&T&F&F&T&F&F&F&T\\ F&F&F&T&F&F&F&F&F&T&T&F \end{array} \]

In the first row, all three wffs true. Therefore, the set is PL-consistent.

Consistency: Example 2

Are \((P\rightarrow Q)\), \((\neg R\vee Q)\), and \((R\wedge\neg Q)\) PL-consistent?

\[ \begin{array}{c c c|c c c|c c c|c c c} P&Q&R&P\rightarrow Q&\neg R\vee Q&R\wedge\neg Q\\ \hline T&T&T&T&T&F\\ T&T&F&T&T&F\\ T&F&T&F&F&T\\ T&F&F&F&T&F\\ F&T&T&T&T&F\\ F&T&F&T&T&F\\ F&F&T&T&F&T\\ F&F&F&T&T&F \end{array} \]

There is no row where all the wffs are \(T\), so the set is PL-inconsistent.

Consistency: Application

• Obvious: “John is tall” and “Mary is tall” are consistent. They can both be true.
• Not so obvious: “If John is tall, then Mary is happy” and “John is not tall or Mary is happy.”

Continued

1. Translation: \(J\rightarrow M\) and \(\neg J\vee M\).
2. Construct the table and check: PL-consistent or PL-inconsistent.

\[ \begin{array}{c c|c c c|c c c c} J&M&J&\rightarrow&M&\neg&J&\vee&M\\ \hline T&T&T&T&T&F&T&T&T\\ T&F&T&F&F&F&T&F&F\\ F&T&F&T&T&T&F&T&T\\ F&F&F&T&F&T&F&T&F \end{array} \]

There is at least one row where both wffs are true. Thus, the sentences are PL-consistent.

Exercise

See book, Ex. 3.34, pp.125-6, #8: \(\neg P\to R, R\to\neg P\)

Equivalence

Equivalence

NoteDefinition

A pair of propositions \(P\) and \(Q\) are equivalent provided whenever \(P\) is true, \(Q\) is true, and whenever \(P\) is false, \(Q\) is false.

NoteDefinition

Members of a set of wffs \(\Gamma\) are PL-equivalent if and only if, for every interpretation, the members have the same truth value: \(v(\gamma_1)=\cdots=v(\gamma_n)\).

If two wffs have matching truth values under every interpretation, then they are equivalent.

Equivalence: A Trivial Example

• \(P\) and \(P\) are equivalent.
• If \(\mathscr{I}(P)=T\), then \(\mathscr{I}(P)=T\).
• If \(\mathscr{I}(P)=F\), then \(\mathscr{I}(P)=F\).

Thus, for every interpretation, whenever \(P\) is true, \(P\) is true, and whenever \(P\) is false, \(P\) is false.

Testing Equivalence

1. Write each wff down in a row.
2. Construct a single truth table.
3. For each row, check whether the truth values match.
4. If they match in every row, the wffs are equivalent.
5. If they do not match in every row, the wffs are not equivalent.

Equivalence: Example

Are \(P\leftrightarrow Q\) and \(P\rightarrow Q\) PL-equivalent?

\[ \begin{array}{c c|c c c|c c c} P&Q&P&\leftrightarrow&Q&P&\rightarrow&Q\\ \hline T&T&T&T&T&T&T&T\\ T&F&T&F&F&T&F&F\\ F&T&F&F&T&F&T&T\\ F&F&F&T&F&F&T&F \end{array} \]

No. They are not PL-equivalent.

Equivalence: Application

Translate the following as \(J\rightarrow M\) and \(\neg J\vee M\):

If John is tall, then Mary is happy.

John is not tall or Mary is happy.

\[ \begin{array}{c c|c c c|c c c} J&M&J&\rightarrow&M&\neg&J&\vee&M\\ \hline T&T&T&T&T&F&T&T\\ T&F&T&F&F&F&T&F\\ F&T&F&T&T&T&F&T\\ F&F&F&T&F&T&F&T \end{array} \]

The wffs are PL-equivalent.

Validity and Semantic Entailment

Validity

NoteDefinition

An argument is valid if and only if it is impossible for the premises to be true and the conclusion false.

NoteDefinition

A set of PL-wffs \(\Gamma\) semantically entails a PL-wff \(\psi\) if and only if there is no interpretation \(\mathscr{I}\) in which all members of \(\Gamma\) are true and \(\psi\) is false.

If \(A\), \(B\), and \(C\) are premises and \(D\) is the conclusion, then \(A,B,C\) semantically entail \(D\) iff there is no interpretation where \(A\), \(B\), and \(C\) are all true and \(D\) is false.

Semantic Entailment

1. We can say that \(\Gamma\) semantically entails \(\psi\), or that \(\psi\) is a semantic consequence of \(\Gamma\).
2. We use the double turnstile, \(\models\), to express entailment.
3. \(\Gamma\models\psi\) says “\(\Gamma\) semantically entails \(\psi\).”
4. If \(\Gamma\) does not semantically entail \(\psi\), we write \(\Gamma\not\models\psi\).

Testing Semantic Entailment

1. Write each wff down in a row.
2. Construct a single truth table.
3. Check for a row where all members of \(\Gamma\) are \(T\) and \(\psi\) is \(F\).
4. If such a row exists, then \(\Gamma\not\models\psi\).
5. If no such row exists, then \(\Gamma\models\psi\).

Semantic Entailment: Example

Does \(P\rightarrow Q\) and \(P\) semantically entail \(Q\)?

\[ \begin{array}{c c|c c c|c|c} P&Q&P&\rightarrow&Q&P&Q\\ \hline T&T&T&T&T&T&T\\ T&F&T&F&F&T&F\\ F&T&F&T&T&F&T\\ F&F&F&T&F&F&F \end{array} \]

Yes. There is no row where \(P\rightarrow Q\) and \(P\) are true and \(Q\) is false. Therefore, \(P\rightarrow Q,P\models Q\).

Semantic Entailment: Example

Does \(P\vee Q\) and \(P\) semantically entail \(Q\)?

\[ \begin{array}{c c|c c c|c|c} P&Q&P&\vee&Q&P&Q\\ \hline T&T&T&T&T&T&T\\ T&F&T&T&F&T&F\\ F&T&F&T&T&F&T\\ F&F&F&F&F&F&F \end{array} \]

No. In row 2, \(P\vee Q\) and \(P\) are true while \(Q\) is false. Therefore, \(P\vee Q,P\not\models Q\).

Semantic Entailment: Application

Translate the argument as \(J\rightarrow M,\neg M\models\neg J\):

If John is tall, then Mary is happy. Mary is not happy. Therefore, John is not tall.

\[ \begin{array}{c c|c c c|c c|c c} J&M&J&\rightarrow&M&\neg&M&\neg&J\\ \hline T&T&T&T&T&F&T&F&T\\ T&F&T&F&F&T&F&F&T\\ F&T&F&T&T&F&T&T&F\\ F&F&F&T&F&T&F&T&F \end{array} \]

Thus, \(J\rightarrow M,\neg M\models\neg J\).

Some Shortcuts

1. If \(\psi\) is a PL-tautology, then \(\Gamma\models\psi\).
2. Since \(\Gamma\not\models\psi\) only when \(\psi\) is false, check rows where \(\psi\) is false.
3. If \(\Gamma\) is PL-inconsistent, then \(\Gamma\models\psi\).
4. Since \(\Gamma\not\models\psi\) only when \(\Gamma\) is PL-consistent, check rows where all wffs in \(\Gamma\) are true.

Limitations of Truth Tables

Problem 1: All valid arguments

• If an argument can be expressed by a truth-functional language (\(PL\)), the table method determines whether it is valid or invalid
• BUT: not every English argument can be represented in a truth-functional language like PL.

Problem 1: Example

• P1. All humans are mortal.
• P2. Socrates is a human.
• C. Socrates is mortal.

Translation

• P1. \(H\)
• P2. \(S\)
• C. \(M\)

Solutions to Problem 1

The fix to this problem is to develop a more powerful logical language.

Problem 2: Exponential increase

For each new propositional letter \(n\), the table grows \(2^n\)

import matplotlib.pyplot as plt

n = list(range(1, 11))
rows = [2**k for k in n]

fig, ax = plt.subplots(figsize=(10, 5))
ax.plot(n, rows, marker="o", linewidth=2.5, color="#2C7FB8")

for letters, count in zip(n, rows):
    ax.annotate(
        f"{count:,}",
        (letters, count),
        xytext=(0, 8),
        textcoords="offset points",
        ha="center",
        fontsize=12,
    )

ax.set_title(r"Truth-table size grows exponentially: $2^n$")
ax.set_xlabel("Number of propositional letters ($n$)")
ax.set_ylabel("Number of rows ($2^n$)")
ax.set_xticks(n)
ax.grid(True, axis="y", alpha=0.3)
ax.spines[["top", "right"]].set_visible(False)
fig.tight_layout()
plt.show()

Solution to Problem 2

Develop a better test.