Using the Rules of Inference
In the previous section, we collected a large sample of commonly occurring logical implications. We briefly touched on why we’d want such a list — to determine whether a given argument is valid. In addition to determining validity, we can also use the rules of inference to make valid deductions from a given list of premises.
In this section, we work through several examples of both use cases.
Determining an Argument’s Validity
Suppose we’re presented with some argument: a list of premises, and a desired conclusion. We can determine if the argument is valid by appealing to the rules of inference.
Because you love live rock music, you decide to purchase front-row tickets for an upcoming rock concert. The tickets are expensive, so you’ll need to save up money working a summer job to purchase them. The problem is that everybody wants front-row seats, so they may be sold out by the time you have enough money.
Consider the propositions
\[ \begin{array}{rl} a\text{: } &\text{You save up enough money to purchase front-row seats.} \\ b\text{: } &\text{There are no more front-row seats available.} \\ c\text{: } &\text{You sit front row at the rock concert.} \end{array} \]and the argument
\[ \begin{array}{l} \neg b \\ \neg b \to a \\ a \to c \\ \hline \therefore c \end{array} \]To determine whether this argument is valid, notice that because we have both $\neg b$ and $\neg b \to a$, we must have $a$ by Modus Ponens. Now, because we have both $a$ and $a \to c$, we also have $c$ by Modus Ponens.
We just reached the desired conclusion $c$ by appealing to Modus Ponens twice, meaning the argument is valid. So, if there are front-row seats available, you’ll be able to save up enough money to sit front row at the rock concert. Awesome!
It seems a bit cumbersome to write out our logic in paragraphs like this. Just like we did when showing two compound propositions were logically equivalent, we can write out a sequence of steps in tabular form.
Luckily for us, another rock concert is happening, which means we need to start saving even more money, hoping front-row seats are still available. Reconsider the argument from the previous example. We can write out the sequence of steps we took there in tabular form:
\[ \begin{array}{lll} \textbf{Step} & \textbf{Proposition} & \textbf{Reason} \\ (1) & \neg b & \text{Premise} \\ (2) & \neg b \to a & \text{Premise} \\ (3) & a & \text{Modus Ponens on Steps (1) and (2)} \\ (4) & a \to c & \text{Premise} \\ (5) & \therefore c & \text{Modus Ponens on Steps (3) and (4)} \end{array} \]Just like before, we reached conclusion $c$ using the rules of inference. We’ll use this tabular form of validating an argument from here on out.
There are many rules of inference, so we may be able to validate a given argument in multiple different ways.
Let’s reconsider the argument once more. Instead of using Modus Ponens twice, we could look at the big list of inference rules from the previous section again. One rule that stands out is the Law of the Syllogism, since we have two implications as premises:
\[ \begin{array}{lll} \textbf{Step} & \textbf{Proposition} & \textbf{Reason} \\ (1) & \neg b \to a & \text{Premise} \\ (2) & a \to c & \text{Premise} \\ (3) & \neg b \to c & \text{Law of the Syllogism on Steps (1) and (2)} \\ (4) & \neg b & \text{Premise} \\ (5) & \therefore c & \text{Modus Ponens on Steps (3) and (4)} \end{array} \]Of course, we arrive yet again at the desired conclusion $c$.
Going forward, when we write out these tabular arguments, we’ll omit the word “Steps,” and just write out which numbered step is being used in a rule of inference — this will save us a bit of writing. It’s also worth pointing out that since some propositions are given as premises, they require no justification beyond noting they’re premises of the argument.
Some arguments require multiple rules of inference to determine validity.
Consider the following argument, with propositions $s$, $t$, $x$, $y$, and $z$:
\[ \begin{array}{l} x \\ x \to y \\ s \lor t \\ t \to \neg y \\ \hline \therefore s \lor z \end{array} \]We could take the following steps to validate this argument:
\[ \begin{array}{lll} \textbf{Step} & \textbf{Proposition} & \textbf{Reason} \\ (1) & x & \text{Premise} \\ (2) & x \to y & \text{Premise} \\ (3) & y & \text{Modus Ponens on (1) and (2)} \\ (4) & t \to \neg y & \text{Premise} \\ (5) & y \to \neg t & \text{Contrapositive of (4): } (t \to \neg y) \Longleftrightarrow (y \to \neg t) \\ (6) & \neg t & \text{Modus Ponens on (3) and (5)} \\ (7) & s \lor t & \text{Premise} \\ (8) & s & \text{Rule of Disjunctive Syllogism on (6) and (7)} \\ (9) & \therefore s \lor z & \text{Rule of Disjunctive Amplification on (8)} \end{array} \]So, we arrive at the desired conclusion $s \lor z$, using a wide variety of rules of inference.
In the previous example, step (5) made use of a logical equivalency between contrapositives. As we work through an argument, we can introduce logically equivalent propositions whenever we want — so we should make use of this as much as possible.
For arbitrary propositions $a$, $b$, $c$, $d$, $e$, and $f$, consider the argument
\[ \begin{array}{l} a \to e \\ e \to (b \land c) \\ \neg c \lor (f \lor \neg d) \\ d \land a \\ \hline \therefore f \end{array} \]This one may require a lot of work, so let’s get started:
\[ \begin{array}{lll} \textbf{Step} & \textbf{Proposition} & \textbf{Reason} \\ (1) & a \to e & \text{Premise} \\ (2) & e \to (b \land c) & \text{Premise} \\ (3) & a \to (b \land c) & \text{Law of the Syllogism on (1) and (2)} \\ (4) & d \land a & \text{Premise} \\ (5) & a & \text{Conjunctive Simplification on (4)} \\ (6) & b \land c & \text{Modus Ponens on (3) and (5)} \\ (7) & c & \text{Conjunctive Simplification on (6)} \\ (8) & \neg c \lor (f \lor \neg d) & \text{Premise} \\ (9) & f \lor \neg d & \text{Disjunctive Syllogism on (7) and (8)} \\ (10) & d & \text{Conjunctive Simplification on (4)} \\ (11) & \therefore f & \text{Disjunctive Syllogism on (9) and (10)} \end{array} \]Making Valid Inferences
When determining whether an argument is valid, we’re given a list of premises and a conclusion, and we write out our justifications in tabular form, with the conclusion as the table’s last entry. Theoretically, we could do the same thing even without being given a conclusion — we just wouldn’t have a “goal” to reach. We could apply whatever rule of inference seems appropriate to the given premises, and to any previous conclusions reached from them.
Consider the propositions
\[ \begin{array}{rl} c\text{: } &\text{I am clever.} \\ \ell\text{: } &\text{I am lucky.} \\ w\text{: } &\text{I win the lottery.} \end{array} \]and the premises $c \lor \ell$, $\neg \ell$, and $\ell \to w$. We’re not given a conclusion, but can we make any inference from these premises?
One conclusion we could easily reach is to use the Rule of Disjunctive Syllogism on the first two premises, giving us conclusion $c$. As such, we know the argument $[(c \lor \ell) \land \neg \ell \land (\ell \to w)] \to c$ is valid.
Note that once you use a rule of inference on a given list of premises, you’re making a valid argument — every intermediate step in the previous section’s longer example produced a valid argument, since each was constructed by means of a rule of inference.
Consider the premises
\[ \begin{array}{l} \text{If the band can't perform their concert, or their t-shirts aren't} \\ \text{available for purchase at the concert, then the after-party will be} \\ \text{cancelled, and you will not purchase front-row seats. If the} \\ \text{after-party is cancelled, then ticket sales will have to be issued} \\ \text{refunds. No refunds were issued.} \end{array} \]We pick out the propositions
\[ \begin{array}{rl} a\text{: } &\text{The band can perform their concert.} \\ t\text{: } &\text{The band's t-shirts are available for purchase.} \\ p\text{: } &\text{The after-party was cancelled.} \\ y\text{: } &\text{You do not buy front-row seats.} \\ r\text{: } &\text{Ticket sales are issued refunds.} \end{array} \]giving us the premises $(\neg a \lor \neg t) \to (p \land y)$, $p \to r$, and $\neg r$. Let’s see what deductions we can make:
\[ \begin{array}{lll} \textbf{Step} & \textbf{Proposition} & \textbf{Reason} \\ (1) & p \to r & \text{Premise} \\ (2) & \neg r & \text{Premise} \\ (3) & \neg p & \text{Modus Tollens on (1) and (2)} \\ (4) & \neg p \lor \neg y & \text{Disjunctive Amplification on (3)} \\ (5) & \neg (p \land y) & \text{DeMorgan's Law on (4)} \\ (6) & (\neg a \lor \neg t) \to (p \land y) & \text{Premise} \\ (7) & \neg (\neg a \lor \neg t) & \text{Modus Tollens on (5) and (6)} \\ (8) & \neg \neg a \land \neg \neg t & \text{DeMorgan's Law on (7)} \\ (9) & a \land t & \text{Law of Double Negation on (8)} \\ (10) & a & \text{Conjunctive Simplification on (9)} \\ (11) & t & \text{Conjunctive Simplification on (9)} \end{array} \]Notice that one of our conclusions was $a$, in step (10). As such, we know that $[((\neg a \lor \neg t) \to (p \land y)) \land (p \to r) \land \neg r] \to a$ is a valid argument — with the given premises, we could deduce that the band performed their concert!
We didn’t stop at that one conclusion, though. Step (11) left us with conclusion $t$, meaning we could also deduce that the band’s t-shirts were available for purchase. Yet another inference we made was $\neg p$, in step (3), meaning the after-party was not cancelled!
Any of the intermediate propositions that weren’t premises are valid inferences from the given premises.
One more strategy we could use is a truth table, to see what combinations of truth values for the propositions yield true premises.
Consider the propositions
\[ \begin{array}{rl} s\text{: } &\text{Johnny had to go to summer school.} \\ j\text{: } &\text{Johnny could work a summer job.} \\ a\text{: } &\text{Johnny could purchase front-row seats at the rock concert.} \end{array} \]and the premises $s$, $s \to \neg j$, and $\neg j \to \neg a$. Constructing a truth table and checking which rows make all three premises true reveals only one combination: $s = 1$, $j = 0$, $a = 0$.
So, we need some combination of these three propositions that yields $1$ under that assignment. One such example is $\neg j$, meaning $[s \land (s \to \neg j) \land (\neg j \to \neg a)] \to \neg j$ is a valid argument. We also have $\neg a = 1$, so $[s \land (s \to \neg j) \land (\neg j \to \neg a)] \to \neg a$ is valid too.
Since $j = 0$ and $\neg a = 1$, we have $j \land \neg a = 0$, meaning $\neg (j \land \neg a) = 1$, and so $\neg j \lor a = 1$ as well. As such, the argument $[s \land (s \to \neg j) \land (\neg j \to \neg a)] \to (\neg j \lor a)$ is also valid.
It’s good practice to try and come up with a sequence of inference rules to reach these kinds of conclusions without going through a truth table.