1.6: Exercises
Let \(p\text{,}\) \(q\text{,}\) and \(r\) represent the following statements.
\begin{align*} p\colon\quad & \text{The game is on.} \\ q\colon\quad & \text{The popcorn is ready.} \\ r\colon\quad & \text{Joe is happy.} \end{align*}
Suppose the following compound statement is true.
If the game is on and the popcorn is ready, then Joe is happy.
However, you just visited Joe's residence room and found that Joe is unhappy even though the game is on. What can you conclude about the popcorn? Use a truth table to justify your answer.
Consider the logical statement \((p \land q_1) \rightarrow (q_1 \lor q_2)\text{.}\) Partially translated, this statement says:
if \(p\) and \(q_1\) are both true, then at least one of \(q_1\) and \(q_2\) is true.
Would you expect this statement to be a tautology? … a contradiction? … neither?
Use a truth table to check.
The wizard Hatty Porrer is studying logic at Cowpimples School for Second-Rate Wizards. As an exercise, he is filling out the truth table for the conditional
\begin{equation*} (r \land s) \rightarrow (r \lor s) \text{.} \end{equation*}
But he forgets what to do for the lines where the “if” part of the conditional evaluates to false, so he only gets this far:
| \(r\) | \(s\) | \(r \land s\) | \(r \lor s\) | \((r \land s) \rightarrow (r \lor s)\) |
| \(T\) | \(T\) | \(T\) | \(T\) | \(T\) |
| \(T\) | \(F\) | \(F\) | \(T\) | ? |
| \(F\) | \(T\) | \(F\) | \(T\) | ? |
| \(F\) | \(F\) | \(F\) | \(F\) | ? |
- Help Hatty out by finishing his homework for him.
- While you were filling out the truth table, Hatty got bored and opened up a portal to a parallel universe. Parallel Hatty is also working on the same truth table, and is stuck at the same spot that normal Hatty was. However, you notice that parallel Hatty's textbook is open to the page with the truth table for the basic conditional \(p \rightarrow q\text{,}\) and it looks as follows.
| \(p\) | \(q\) | \(p \rightarrow q\) |
| \(T\) | \(T\) | \(T\) |
| \(T\) | \(F\) | \(F\) |
| \(F\) | \(T\) | \(F\) |
| \(F\) | \(F\) | \(T\) |
Finish parallel Hatty's homework exercise. Make sure parallel Hatty's instructor will like the result!
- While you were finishing parallel Hatty's homework, Hatty got bored again and opened up a portal to another parallel universe! Parallel Hatty number two is also working on the same truth table, and is stuck at the same spot that the previous two Hattys were. This time, however, parallel Hatty number two's textbook says that the truth table for the basic conditional \(p \rightarrow q\) is as follows.
| \(p\) | \(q\) | \(p \rightarrow q\) |
| \(T\) | \(T\) | \(T\) |
| \(T\) | \(F\) | \(F\) |
| \(F\) | \(T\) | \(T\) |
| \(F\) | \(F\) | \(F\) |
Finish parallel Hatty number two's homework exercise. Make sure parallel Hatty number two's instructor will approve!
- You'll never believe what happened while you were finishing parallel Hatty number two's homework! Yep, Hatty got bored again and opened up a portal to a third parallel universe. Parallel Hatty number three is also working on the same truth table, and is stuck at the same spot that the previous three Hattys were. The truth table for the basic conditional \(p \rightarrow q\) is different in parallel Hatty number three's universe, yet again.
| \(p\) | \(q\) | \(p \rightarrow q\) |
| \(T\) | \(T\) | \(T\) |
| \(T\) | \(F\) | \(F\) |
| \(F\) | \(T\) | \(F\) |
| \(F\) | \(F\) | \(F\) |
Finish parallel Hatty number three's homework exercise. Make sure parallel Hatty number three's instructor will give him full marks!
- OK, so what the heck is the point of all this? The statement \((r \land s) \rightarrow (r \lor s)\) could be read as:
If \(r\) and \(s\) are both true statements, then at least one of \(r\) and \(s\) is a true statement.
This conditional statement seems “obviously true”. Based on this example, what do you think of each parallel universe's system of logic compared to our own?
Suppose \(A,E,U\) are logical statements such that \(U\) is a tautology and \(E\) is a contradiction.
- Show that \(A \lor U\) is always a tautology.
- Show that \(A \land E\) is always a contradiction.
Suppose that \(A\text{,}\) \(B\text{,}\) and \(C\) are logical statements such that \(A \Rightarrow B\) and \(B \Rightarrow C\text{.}\) Must \(A \Rightarrow C\text{?}\)