Always True, Always False

Let's examine the proposition $p \rightarrow (p \lor q)$. We'll start with a blank truth table.

We start by filling out the column for $p \lor q$:

Now that we have the truth values for propositions $p$ and $p \lor q$, it is easy to calculate the truth values for $p \rightarrow (p \lor q)$:

Now we remove the extraneous column for $p \lor q$ to get the final truth table:

All of the values in the final column are $1$! This means that, no matter what truth values $p$ and $q$ have, the expression $p \rightarrow (p \lor q)$ is always $\text{true}$. $$~$$ Translating the proposition $p \rightarrow (p \lor q)$ into English yields something like this: $$\text{If p, then p or q.}$$ Of course if we start with proposition $p$, then we must also have $p \lor q$ as well because $p$ is included in the conclusion. For example, suppose we had that \[ \begin{align*} p &: \text{Phil opens up a new savings account.} \\ q &: \text{Phil earns 4% interest every year, compounded monthly.} \end{align*} \] Then the proposition $p \rightarrow (p \lor q)$ can be translated as follows: \[ \begin{align*} &\text{If Phil opens up a new savings account, then} \\ &\text{either Phil opens up a new savings account, or } \\ &\text{he will earn 4% interest every year compounded monthly, } \\ &\text{or both.} \end{align*} \] $$~$$ Of course if Phil opens a new savings account, then he will either open up a new savings account (which he just supposedly did), or he will earn 4% interest compounded monthly, or both.

Suppose we wanted to examine the proposition $p \land (\neg p \land q)$. Naturally, we start with an empty truth table.

We propositions $\neg p$ and $\neg p \land q$ are relatively easy to calculate. Once those propositions' columns are filled out, then filling out the final column becomes straightforward:

Finally, we'll remove the columns for any propositions we don't care about:

Every value in the last column is $0$, meaning no matter what combination of truth values we have for $p$ and $q$, the proposition $p \land (\neg p \land q)$ will always be $\text{false}$. Then again, perhaps we shouldn't be surprised. After all, how can both $p$ and $\neg p$ be true? To make this situation more concrete, take another look at the propositions we used in Example 1.3.1: \[ \begin{align*} p &: \text{Phil opens up a new savings account.} \\ q &: \text{Phil earns 4% interest every year, compounded monthly.} \end{align*} \] With these propositions, we can translate $p \land (\neg p \land q)$ into something like the following: \[ \begin{align*} &\text{Phil opens up a new savings account, and} \\ &\text{he does not open up a new savings account, and} \\ &\text{he will earn 4% interest every year compounded monthly.} \end{align*} \] How can Phil open up a new savings account while simultaneously not opening up a new savings account? This is an impossible scenario.

This time, let's look at the compound proposition $(p \lor q) \land r$ that is made up of three constituent propositions $p, q, r$:

After filling out the column for $p \lor q$, the last column becomes easy to fill out:

Here, we see there are only three rows where $(p \lor q) \land r = 1$. We'll highlight those rows in green after stripping out the extraneous $p \lor q$ column:

Based on the above truth table, we see that $(p \lor q) \land r = 1$ when we have the following: \[ \begin{align*} p = 0,\ q = 1,\ r = 1 \\ p = 1,\ q = 0,\ r = 1 \\ p = 1,\ q = 1,\ r = 1 \end{align*} \]

Let's take another look at the truth table for the proposition $p \rightarrow (p \lor q)$:

As discussed previously, all truth value assignments for $p$ and $q$ result in a truth value of $1$ for $p \rightarrow (p \lor q)$. Thus, $p \rightarrow (p \lor q)$ is a tautology, and we can write $$\bigl(p \rightarrow (p \lor q)\bigr) = T_0$$

Consider the proposition $\neg p \land (p \land q)$, which is slightly different than the proposition given in Example 1.3.2. Here is it's truth table:

Here, there is no way to choose truth values for $p$ and $q$ to make $\neg p \land (p \land q) = 1$. As such, we have that $$\bigl(\neg p \land (p \land q)\bigr) = F_0$$

As demonstrated in Example 1.3.3, since there exists truth values for $p, q, r$ such that $\bigl((p \lor q) \land r\bigr) = 1$, we have that $\bigl((p \lor q) \land r\bigr)$ is satisfiable.

From Example 1.3.4, we saw that there exists truth values for $p, q$ such that $\bigl(p \rightarrow (p \lor q)\bigr) = 1$, meaning $p \rightarrow (p \lor q)$ is satisfiable.

From Example 1.3.5, we saw that $\bigl(\neg p \land (p \land q)\bigr) = F_0$, meaning $\neg p \land (p \land q)$ is unsatisfiable.