Logical Implications

Suppose we have $p \to q$ where

\[ \begin{array}{rl} p\text{: } &\text{Alyssa studies for her Chemistry exam.} \\ q\text{: } &\text{Alyssa gets an A on her Chemistry exam.} \end{array} \]

What can we deduce if we know $\neg p$ is true (meaning $p$ is false)? In this scenario, Alyssa doesn’t study for her Chemistry exam. Does this mean she doesn’t get an A on her Chemistry exam?

Not necessarily. Remember, all we know is that if she studies, she’ll get an A — we weren’t told $\neg p \to \neg q$. Perhaps the exam is easy enough that Alyssa doesn’t feel the need to study. Perhaps the exam is difficult, but Alyssa is comfortable enough with the material to work out correct answers with a little thought. On the other hand, the exam could be extremely difficult for Alyssa, so maybe she doesn’t get an A. All we can conclude is that she didn’t study — we can’t determine whether we have $q$ or $\neg q$.

What if we know we have $p$ instead of $\neg p$? Since we know that if Alyssa studies, she’ll get an A, knowing that $p$ happened tells us that $q$ happened as well. (Note that since we’re asserting $p \to q$ is true, we can’t have $p \to \neg q$.)

Our conclusion hinges entirely on the truth value of $p$: if $\neg p$ happened, we don’t get any additional conclusions, but if $p$ happened, we immediately know $q$ happens too. This is exactly the behavior of the disjunction $\neg p \lor q$ — which is precisely why $p \to q \Longleftrightarrow \neg p \lor q$ in the first place.

Consider two propositions $a$ and $b$, where

\[ \begin{array}{rl} a\text{: } &\text{Alvarez hauls up a red king crab pot.} \\ b\text{: } &\text{Alvarez has to report the catch to the harbormaster.} \end{array} \]

Let’s compare two conjunctions built from these propositions and the implication $a \to b$: $a \land (a \to b)$, and $b \land (a \to b)$.

Suppose $a \land (a \to b)$ is true. Then $a$ is true — Alvarez hauls up a red king crab pot — and $a \to b$ is true as well. Since $a$ is true and $a \to b$ is true, we can deduce $b$ must be true too — Alvarez has to report the catch. So $a \land (a \to b)$ being true pins down both $a$ and $b$.

Now suppose $b \land (a \to b)$ is true instead. Then $b$ is true — Alvarez has to report the catch. But once $b$ is true, the implication $a \to b$ is automatically true too, no matter what $a$ happens to be — an implication with a true conclusion can never be false. So $b \land (a \to b)$ being true doesn’t actually tell us anything about $a$; it only ever tells us that $b$ is true.

The truth table below confirms that $[a \land (a \to b)] \to b$ is a tautology:

Every row in the $[a \land (a \to b)] \to b$ column is a $1$, so $a \land (a \to b)$ always forces $b$ to be true.

Now compare that against the truth table for $[b \land (a \to b)] \to a$:

Here, the $[b \land (a \to b)] \to a$ column has a single $0$, so this implication is not a tautology — $b \land (a \to b)$ being true doesn’t let us conclude $a$.