In order to analyze arguments, we need to understand not only compound statements, but also the relationships between compound statements.
That is, we seek to understand the how the truth of a statement, for example $$p \rightarrow q$$ varies in relation to another statement, for example $$p \wedge \sim q.$$
Types of Statements: Tautology
Defintion: A statement which is true for all combination of truth values is called a tautology.
Silly Example: $p \vee \sim p$
We'll get to more serious examples a little later
Types of Statements: Self-Contradictions
Defintion: A statement which is false for all combination of truth values is called a self-contradiction.
Yet Another Silly Example: $p \wedge \sim p$
We'll get to more serious examples right now!
Example
Use a truth table to decide if the statement $(p \vee q) \wedge (\sim p \rightarrow q)$ is a tautology, self-contradiction, or neither.
Example
Use a truth table to decide if the statement $(p \wedge \sim q) \wedge \sim p$ is a tautology, self-contradiction, or neither.
Example
Use a truth table to decide if the statement $(p \rightarrow q) \vee \sim q$ is a tautology, self-contradiction, or neither.
Logically Equivalent Statements
Definition: Two statements are logically equivalent if they always have the same truth value.
Silly Example: $p$ and $\sim \sim p$ are logically equivalent. This is simply the double negative.
We'll get to a serious example right now. And yes, it's serious!
Logically Equivalent Statements
Use a truth table to show that the statements $\sim (p \rightarrow q)$ and $p \wedge \sim q$ are logically equivalent.
Notational Note: We may write the above logical equivalence as $$\sim (p \rightarrow q) \equiv p \wedge \sim q.$$
Pulling It Back Down To Earth
Ok, so we've proven that $\sim (p \rightarrow q) \equiv p \wedge \sim q.$
Sometimes it's good to use an example to make sure our heads aren't too far off in the clouds.
Let $p$ be the statement Mr. Holt is hungry, and $q$ be the statement Mr. Holt is grumpy. With a little bit of thought, you can see that both of the following statements are negations of each other.
$p \rightarrow q$: If Mr. Holt is hungry, then Mr. Holt is grumpy.
$p \wedge \sim q$: Mr. Holt is hungry and Mr. Holt is not grumpy.
De Morgan's Laws for Logic
For any statements $p$ and $q,$ the following hold: $$\sim (p \vee q) \equiv \sim p \wedge \sim q$$ $$\sim (p \wedge q) \equiv \sim p \vee \sim q$$
De Morgan's Laws for Logic
Use De Morgan's Laws to show that $p \rightarrow q \equiv \sim p \vee q.$
Negations of Conjunction and Disjunction
De Morgan's Laws are also very handy for negating and statements and or statements.
Example: Find the negation of the statement: Mr. Holt is hungry and I won't take an umbrella.
Mr. Holt is not hungry or I will take an umbrella.
Example: Find the negation of the statement: Linda Lou is serious about her studies or Billy Bob is not serious about his studies.
Linda Lou is not serious about her studies and Billy Bob is serious about
his studies.
Variations on Conditional Statements
Conditional statements play a large role in arguing a point. Again, in order to separate good arguments from bad ones, we will need to learn the following conditional forms. $$ \begin{array}{|l|l|l|} \hline \mbox{Name} & \mbox{Symbolic Statement} & \mbox{Word Statement} \\\hline \mbox{Conditional} & p \rightarrow q & \mbox{If $p,$ then $q.$} \\ \mbox{Converse} & q \rightarrow p & \mbox{If $q,$ then $p.$} \\ \mbox{Contrapositive} & \sim q \rightarrow \sim p & \mbox{If not $q,$ then not $p.$} \\ \mbox{Inverse} & \sim p \rightarrow \sim q & \mbox{If not $p,$ then not $q.$} \\ \hline \end{array} $$
Example
Let $p$ be the statement Mr. Holt is hungry, and $q$ be the statement Mr. Holt is grumpy.
Write the conditional, converse, contrapositive, and inverse statements.
Conditional: If Mr. Holt is hungry, then Mr. Holt is grumpy.
Converse: If Mr. Holt is grumpy, then Mr. Holt is hungry.
Contrapositive: If Mr. Holt isn't grumpy, then Mr. Holt isn't hungry.
Inverse: If Mr. Holt isn't hungry, then Mr. Holt isn't grumpy.
Converse: If Mr. Holt is grumpy, then Mr. Holt is hungry.
Contrapositive: If Mr. Holt isn't grumpy, then Mr. Holt isn't hungry.
Inverse: If Mr. Holt isn't hungry, then Mr. Holt isn't grumpy.
What's Ahead: As we shall see in the next section, many a bad argument comes from assuming that all of these variations on the conditional are logically equivalent.
Hopefully, the above examples illustrate that this is not the case. For example, the conditional and the converse are not logically equivalent:
Conditional: If Billy Bob gets caught making moonshine, then he will go to jail.
Converse: Billy Bob went to jail, so he must have been making moonshine.
One of the most common mistakes in argumentation is assuming the converse ($q \rightarrow p$) is logically equivalent to the conditional ($p \rightarrow q$). In the next section we'll see others.
Truth Tables for Conditional, Converse, Contrapositive, and Inverse $$ \begin{array}{|c|c|c|c|c|c|c|c|} \hline p & q & \sim p & \sim q & \color{blue}{p \rightarrow q} & \color{red}{q \rightarrow p} & \color{blue}{\sim q \rightarrow \sim p} & \color{red}{\sim p \rightarrow \sim q} \\ \hline T & T & F & F & \color{blue}{T} & \color{red}{T} & \color{blue}{T} & \color{red}{T}\\ T & F & F & T & \color{blue}{F} & \color{red}{T} & \color{blue}{F} & \color{red}{T}\\ F & T & T & F & \color{blue}{T} & \color{red}{F} & \color{blue}{T} & \color{red}{F}\\ F & F & T & T & \color{blue}{T} & \color{red}{T} & \color{blue}{T} & \color{red}{T}\\ \hline \end{array} $$ From the above, we see that the conditional and the contrapositive are logically equivalent $$\color{blue}{p \rightarrow q \equiv \sim q \rightarrow \sim p}$$ that the converse and the inverse are logically equivalent $$\color{red}{q \rightarrow p \equiv \sim p \rightarrow \sim q}$$
Mathematical Aside Stop Taking Notes! The following will NOT be on the exam.
Recall that sets and logic are strongly connected. In fact, we can prove De Morgan's laws for sets using De Morgan's laws for logic: $$ \begin{array}{l|l} (A \cup B)' & (A \cap B)'\\ =\{x| \sim (x \in A \vee x \in B) \} & =\{x| \sim (x \in A \wedge x \in B) \} \\ =\{x| \sim (x \in A) \wedge \sim (x \in B) \} & =\{x| \sim (x \in A) \vee \sim (x \in B) \} \\ =A' \cap B' & =A' \cup B' \end{array} $$