As described in the course introduction, this is a course first and foremost about systematic deductive reasoning. There are two major learning goals: to learn to systematically solve problems, and to learn to systematically evaluate solutions to problems. I can not stress enough that these are not only two of the most generalizably important skills in computer science, but will also become increasingly important as generative AI becomes better at writing code.
In the first unit, much of the content will focus on the second learning goal - evaluating solutions to problems. This may appear counterintuitive at first glance - why learn to evaluate things we haven't learned to create? - but I think this is actually a much more natural way to learn. Every idiom in the world tells us that the key to success is learning from mistakes (What doesn't kill you makes you stronger! Our greatest glory is not in falling, but in rising every time we fall! etc...) but how can we learn from our mistakes if we can't even tell whether they're wrong?
The ability to distinguish true from true-sounding is increasingly important as our society progresses. As I'm sure you know, we will determine whether things are true by writing proofs.
The backbone of any proof is logic. In our course and in this field, it will be used as our model for determining truth. Practically speaking, a brief encounter with logic will give you a foundation for parsing, understanding, and finally proving mathematical statements.
We will begin with propositional logic and expand from there. The basic element of propositional logic is the proposition. Propositions simply model statements that are either true or false.
A proposition is a statement that is either true or false.
The following sentences are propositions:
The following are not propositions:
The upshot is that we'll be very permissive about what counts as a proposition. As long as we can reasonably interpret a statement as an assertion about something that can be true or false, it's a proposition.
If propositional logic stopped there, it wouldn't be very interesting. But we can already notice that simply modeling statements as true and false suggests some intuitive relationships. For instance, we don't expect that the first two examples can both be true (since Illinois only has one capitol), and the third example is clearly related somehow to the first two.
In propositional logic, we represent propositions symbolically by formulas. Formulas are made up of propositional variables, parentheses, and connectives. Propositional variables, usually lowercase roman letters like $p,q,r,\dots$, represent atomic propositions and connectives allow us to build compound propositions by joining one or more propositions together. Parentheses are used to denote the ordering that we should interpret the connectives in. Hypothetically, we would be able to express every proposition that shows up in this course symbolically, but we won't go that far.
There are four basic logical connectives, called Boolean connectives, after George Boole who defined them in 1854. We will define and consider each one.
For a proposition $p$, $\neg p$ is called negation of $p$, and is pronounced "not $p$". The proposition $\neg p$ is defined to be true when $p$ is false and false when $p$ is true.
For instance, $\neg q$ is the proposition "I cannot take two weeks off". Note that negation is a unary connective, in that it only applies to one formula, while all the other connectives are binary. Observe that because of this, we tend to express atomic propositions positively.
The unary connective $\neg$ is defined for a propositional formula $p$ by
| $p$ | $(\neg p)$ |
|---|---|
| $T$ | $F$ |
| $F$ | $T$ |
This is our first example of a truth table. To construct it, we listed the possible truth values for $p$, and then for each of those we listed $(\neg p)$ in the corresponding row.
For propositions $p$ and $q$, $p \wedge q$ is the conjunction of $p$ and $q$. It is pronounced "$p$ and $q$". The proposition $p \wedge q$ is defined to be true if both $p$ and $q$ are true, and false otherwise.
This connective expresses the idea that both $p$ and $q$ are true. For instance, $q \wedge r$ is the proposition "I can take two weeks off and I can fly to Tokyo".
For propositions $p$ and $q$, $p \vee q$ is the disjunction of $p$ and $q$. It is pronounced "$p$ or $q$". The proposition $p \vee q$ is defined to be true $p$ is true or $q$ is true (or both), and false otherwise.
For example, $q \vee r$ is the proposition "I can take two weeks off or I can fly to Tokyo". One tricky thing to note with English is that many times we use "or" to mean "exclusive or". For instance, when you are asked whether you prefer beef or chicken, the expectation is that you may only choose one and not both. This logical connective allows for both $p$ and $q$ to be true, which corresponds to something like "and/or" in English.
The binary connectives $\wedge$ and $\vee$ are defined for propositional formulas $p$ and $q$ by
| $p$ | $q$ | $\quad$ | $p \wedge q$ | $p \vee q$ |
|---|---|---|---|---|
| $T$ | $T$ | $T$ | $T$ | |
| $T$ | $F$ | $F$ | $T$ | |
| $F$ | $T$ | $F$ | $T$ | |
| $F$ | $F$ | $F$ | $F$ |
So far the connectives have been mostly natural, up to some quibbling over the meaning of "or". But propositional logic really springs to life we speak of assertions implying each other. Essentially every mathematical statement asserts that, under some conditions, one can draw certain conclusions. This motivates the following definition.
For propositions $p$ and $q$, the proposition $p \rightarrow q$ is called implication and is pronounced "if $p$, then $q$". We call $p$ the hypothesis and $q$ the conclusion.
That definition leaves much to be desired. How exactly can we construct a truth table for the $\rightarrow$ connective? Some rows of the table are relatively easy:
| $p$ | $q$ | $\quad$ | $p \rightarrow q$ |
|---|---|---|---|
| $T$ | $T$ | $T$ | |
| $T$ | $F$ | $F$ | |
| $F$ | $T$ | ? | |
| $F$ | $F$ | ? |
Even those rows could be debated, but we're ultimately building up to the language that mathematicians use, so that will be our guide. And in their usage, $p\rightarrow q$ does not mean "$p$ causes $q$ to be true", in the way that sunshine and rain cause grass to grow. The meaning of $p\rightarrow q$ is closer to "if $p$ is true, then $q$ is also true". So if $p$ is true and $q$ is true, then $p\rightarrow q$ is declared to be true. Similarly, if $p$ is true but $q$ is false, then $p\rightarrow q$ is false.
But what about the remaining rows? If $p$ is false, then it's a funny question. In fact, we saw an example of this situation above: If Chicago is the capitol of Illinois, then milkshakes are free. (Chicago is not the capitol of Illinois!) There are a few ways to wrap your head around what the value should be here - the primary one being the vacuous truth: the value of $q$ does not matter if $p$ is false, so for any $q$ the statement $p \rightarrow q$ should hold. Or, we can think of it as saying that if our assumptions are false then we can conclude anything. Regardless, we'll see later that it makes notation much more straightforward in the future if we assume that both of those rows should be filled with $T$, giving the truth table:
| $p$ | $q$ | $\quad$ | $p \rightarrow q$ |
|---|---|---|---|
| $T$ | $T$ | $T$ | |
| $T$ | $F$ | $F$ | |
| $F$ | $T$ | $T$ | |
| $F$ | $F$ | $T$ |
One could certainly fairly object to this, but we'll see in a few lectures that mathematical statements are ultimately a lot more natural if we use this definition.
We can string together propositions using the connectives $\neg, \wedge,\vee,\rightarrow,\leftrightarrow$ to model more complex assertions. As with arithmetic, we need to be careful about the order of operations.
An example of a more complex proposition we'd like to model is "If it is either Wednesday and it is not raining, then I buy a milkshake." This could be modeled as $(p \vee (\neg q)) \rightarrow r$, where $p =$ "It is Wednesday." and $q =$ "It is raining." and $r =$ "I will be a milkshake."
In order to avoid being overwhelmed by parentheses, we introduce an order of precedence amongst the connectives: From highest to lowest, we take them in the order $\neg, \wedge,\vee$ and let $\rightarrow,\leftrightarrow$ have equal precedence (and thus ties are broken by reading them left-to-right). So, for example, $\neg p \wedge q$ means $(\neg p) \wedge q$ and not $\neg (p \wedge q)$ because $\neg$ has higher precedence than $\wedge$. Something like $p\rightarrow q \leftrightarrow r$ will be avoided, because we have to assign precedence from left to right, meaning it is $(p\rightarrow q) \leftrightarrow r$ and not $p\rightarrow (q \leftrightarrow r)$. In these cases we'll try to use parentheses always.
Arbitrarily complex propositions are allowed; For instance we could consider $((p\wedge q)\rightarrow r) \leftrightarrow (r \rightarrow q)$ or other even larger combinations of connectives.
To compute the truth table of a compound proposition, we proceed by computing the truth tables of its components and iteratively put them together, as in the following example.
Let's compute the truth table of $\neg p\wedge q$. We start by computing the truth table of $\neg p$ as before:
| $p$ | $q$ | $\quad$ | $\neg p$ |
|---|---|---|---|
| $T$ | $T$ | $F$ | |
| $T$ | $F$ | $F$ | |
| $F$ | $T$ | $T$ | |
| $F$ | $F$ | $T$ |
Now we use this column, along with the existing column for $p$, the compute the truth table we are after. We simply extend the table with another column:
| $p$ | $q$ | $\quad$ | $\neg p$ | $\quad$ | $\neg p\wedge q$ |
|---|---|---|---|---|---|
| $T$ | $T$ | $F$ | $F$ | ||
| $T$ | $F$ | $F$ | $F$ | ||
| $F$ | $T$ | $T$ | $T$ | ||
| $F$ | $F$ | $T$ | $F$ |
Let's try a slightly harder one to demonstrate that the negation does not naturally distribute: $\neg(p \rightarrow q) \neq \neg p \rightarrow \neg q$.
| $p$ | $q$ | $\quad$ | $\neg p$ | $\neg q$ |
|---|---|---|---|---|
| $T$ | $T$ | $F$ | $F$ | |
| $T$ | $F$ | $F$ | $T$ | |
| $F$ | $T$ | $T$ | $F$ | |
| $F$ | $F$ | $T$ | $T$ |
Again, we extend some columns and fill in the rows using our definitions:
| $p$ | $q$ | $\quad$ | $\neg p$ | $\neg q$ | $\quad$ | $\neg p \rightarrow \neg q$ | $\quad$ | $p \rightarrow q$ | $\neg(p \rightarrow q)$ |
|---|---|---|---|---|---|---|---|---|---|
| $T$ | $T$ | $F$ | $F$ | $T$ | $T$ | $F$ | |||
| $T$ | $F$ | $F$ | $T$ | $T$ | $F$ | $T$ | |||
| $F$ | $T$ | $T$ | $F$ | $F$ | $T$ | $F$ | |||
| $F$ | $F$ | $T$ | $T$ | $T$ | $T$ | $F$ |
So we can see that there are disagreeing values between $\neg(p \rightarrow q)$ and $\neg p \rightarrow \neg q$, so they are not logically equivalent (see below).
Once you do a few of these, they become a simple matter of robotically composing together truth tables until you have the one you're after.
Here's a more complicated example, given in one table, for the proposition $((p \wedge q) \rightarrow r) = (p \rightarrow (q \rightarrow r))$. We can see how even the equals is built in.
| $p$ | $q$ | $r$ | $p \wedge q$ | $(p \wedge q) \to r$ | $q \to r$ | $p \to (q \to r)$ | $(p \wedge q) \to r = p \to (q \to r)$ |
|---|---|---|---|---|---|---|---|
| $T$ | $T$ | $T$ | $T$ | $T$ | $T$ | $T$ | $T$ |
| $T$ | $T$ | $F$ | $T$ | $F$ | $F$ | $F$ | $T$ |
| $T$ | $F$ | $T$ | $F$ | $T$ | $T$ | $T$ | $T$ |
| $T$ | $F$ | $F$ | $F$ | $T$ | $T$ | $T$ | $T$ |
| $F$ | $T$ | $T$ | $F$ | $T$ | $T$ | $T$ | $T$ |
| $F$ | $T$ | $F$ | $F$ | $T$ | $F$ | $T$ | $T$ |
| $F$ | $F$ | $T$ | $F$ | $T$ | $T$ | $T$ | $T$ |
| $F$ | $F$ | $F$ | $F$ | $T$ | $T$ | $T$ | $T$ |
Note that since we had three atomic propositions $p,q,r$ we needed eight rows in our table this time. Also note that, curiously, this proposition is always true.
Consider the proposition:
Hina knows Python if she is a CS major and took 141, and she is a CS major or did not take 141.
A mathematical statement of this form would be rejected outright as too complicated; If you find yourself twisting around words like this, it's a sign that you should refactor your claims into something plainer. But let's try to parse it, just for the challenge. The atomic propositions are
Next, if you look at the statement, it breaks apart into two independent pieces at the comma, joined by a conjunction. The first piece (Hina knows Python if she is a CS major and took 141,) is $(q \wedge r) \rightarrow p$; Note that $q$ and $r$ come second in the English but are on the left-hand side in logical notation. The second piece (she is a CS major or did not take 141) is $q \vee \neg r$.
Putting that all together gives $((q \wedge r) \rightarrow p) \wedge (q \vee \neg r)$. Quite a mouthful.Consider the proposition:
I'm having pie, unless I'm having a milkshake.
The problem here is with "unless". Take $p=$"I'm having pie" and $q=$"I'm having a milkshake". The question is: What does "$p$ unless $q$" mean, logically?
The common interpretation is $\neg q \rightarrow p$, that is, "If I'm not having a milkshake, then I'm having pie.". You can justify this to yourself by writing down a truth table directly from the sentence and comparing it to $\neg q \rightarrow p$ (though the case where $p$ and $q$ are both false is still awkward; it's similar to the situation with $\rightarrow$). In any case, you may see "unless" from time to time, but it is usually best to avoid it.
Two propositions $p,q$ are said to be equivalent if they have the same truth tables. We write $p \equiv q$ to indicate that $p$ and $q$ are equivalent.
The definition is simple because no matter how hairy two propositions are, you can always derive their truth tables mechanistically and then compare them. Why it should be useful will hopefully be apparent later. At a high level, when you're asked to prove a mathematical statement is true, a common trick is to translate it to a proposition $p$, manipulate it into an equivalent proposition $q$, and then prove the equivalent proposition instead.
The rest of this section records a set of equivalences that are "universal" in a sense that logicians study: Any equivalence can be explained by applying a sequence of these equivalences. For our purposes, we present these equivalences to draw attention to some manipulations that you might be applying without even realizing it.
You can check the following equivalences by computing truth tables. These don't need to be memorized, but are worth trying to understand.
In addition to the truth tables, you can intuit why these equivalences should hold. For example, in the first distributed law, if $p$ is true then both sides are true; If $p$ is false, then both sides are true exactly when $q$ and $r$ are both true.
Here are two more famous equivalences:
These equivalences are also fairly intuitive. In fact, you've probably
thought through them at some point while programming. For example,
the code
if( !((x>0) and (y==1)) ) {
// do something
}
if( (x<=0) or (y!=1) ) {
// do something
}
This is another form of De Morgan's Laws.
Below let us write $T$ as a placeholder for any proposition
that is a tautology (definitionally true, e.g. $p \vee \neg p$), similarly $F$ for a contradiction
(definitionally false, e.g. $p \wedge \neg p$).
Verifying these is a good exercise (but don't bother memorizing them
for this course).
Of all the equivalences in this
section, this one is probably the least familiar:
Let's look at the truth tables for these:
| $p$ | $q$ | $\quad$ | $p \rightarrow q$ | $\neg p$ | $\neg p \vee q$ |
|---|---|---|---|---|---|
| $T$ | $T$ | $T$ | $F$ | $T$ | |
| $T$ | $F$ | $F$ | $F$ | $F$ | |
| $F$ | $T$ | $T$ | $T$ | $T$ | |
| $F$ | $F$ | $T$ | $T$ | $T$ |
The truth tables match, so they're definitely equivalent. But why? One way to understand it is that when we assert $p\rightarrow q$, if $p$ is not true, then we're done and it holds. But if $p$ is true, then we need $q$ to hold; This is exactly what the right-hand side of the equivalence says.
Here is a particularly useful equivalence:
$$\begin{aligned}p\rightarrow q \equiv \neg q \rightarrow \neg p.\end{aligned}$$The contraposition of an implication $p \rightarrow q$ is defined as $\neg q \rightarrow \neg p$, and it turns out that these are equivalent! You'll check in the homework that this is true. This isn't true for $q \rightarrow p$ (the converse) or $\neg p \rightarrow \neg q$ (the inverse). Later we'll learn about "switching to the contrapositive" as a very useful trick in finding proofs.
Now, one thing you might notice is that propositions as we've defined them don't quite let us say everything we may want to. For instance, if I wanted to talk about some property about integers, I can certainly define propositions for them. However, there's no way to relate the various propositions about integers to the fact that we're talking about the same kinds of objects. For that, we need to expand our language a bit further. The extension that we'll be talking about is called predicate or first-order logic.
First, we need to determine a domain of discourse, which is the set to which all of our objects belong. This can be something like the natural numbers $\mathbb N$ or the integers $\mathbb Z$ or matrices or functions or graphs and so on. There's also nothing stopping us from making our domain the set of all dogs, if what we really wanted to do was express statements about dogs, or other "real-world" types of objects. This occurs less in math but more in places like database theory.
This brings us to an important thing to keep in mind: statements that may be true in one domain will not necessarily be true in another domain. Obviously, statements that we make about integers will not necessarily hold true for dogs, but it is important to remember that the same could be said for integers versus natural numbers.
We define the following domains/sets:
Note that in this course, we consider 0 to be a natural number. Not everyone agrees! This is mostly to simplify some notation later on.
Then, we want to express properties and relationships about the objects in our domain. Specifically, we may want to know whether a property is true for a specific object in our domain. At first glance we may try the proposition $p$ to be defined as "$x$ is an even number," but we run into a problem - this isn't a proposition because its truth value depends on $x$! For this, we need to define predicates.
For example, if we define $E$ to be a predicate that expresses the evenness property, then we can say that $E(x)$ means "$x$ is an even number". We would then have $E(12000)$ is true while $E(765)$ is false.
The less-than relation is another example of a predicate, although because it's pretty important, we've given it its own symbol and usage. We can define a predicate $L(x,y)$ to mean "$x$ is less than $y$" or "$x \lt y$". Then we can say that $L(3,30)$ is true and $L(30,3)$ is false.
In the previous examples, almost instinctively, we used both concrete objects and variables with our predicates. When we use specific objects with predicates, our statements are not all that different from propositions. However, variables are considered placeholders for objects in the domain and it's the use of variables, together with quantifiers, that gives us more expressive power than propositional logic.
The symbol $\forall$ is called the universal quantifier and is read "for all". The symbol $\exists$ is called the existential quantifier and is read "there exists".
Basically, we use $\forall$ when we want to make a statement about all objects in our domain and we use $\exists$ when we want to say that there is some object out there in our domain that has a particular property.
We can use predicates and quantifiers together with the logical connectives we introduced with propositional logic to say things like $\forall x (E(x) \rightarrow \neg E(x+1))$. If we take our domain to be $\mathbb Z$, this says that for every integer $x$, if $x$ is even, then $x+1$ is not even.
Let's fix our domain to be $\mathbb N$. Recalling our less than predicate $L$ from above, we can consider the following proposition:
$$\forall x \exists y L(x,y).$$This reads "for every natural number $x$, there exists a natural number $y$ such that $x \lt y$". This statement turns out to be true: every natural number does have some other natural number that it's less than. Now, let's consider the following similar statement, where the order of the quantifiers is reversed:
$$\exists y \forall x L(x,y).$$Here, we're saying that there is a natural number $y$ for which every other natural number $x$ is less than it. In other words, this says that there is a "biggest" natural number, which is obviously not true. As we can see, the order of the quantifiers matters.
Remember that the domain is important. For instance, if we are working in $\mathbb N$ and we modify the above example slightly to $\forall x \exists y L(y,x)$, we have the statement that "for every natural number $x$, there is some natural number $y$ that is less than $x$". This statement is false if we consider $x = 0$. However, this statement is true if our domain is $\mathbb Z$ (for every integer $x$, there is some integer $y$ that is less than $x$). ∎
Something we have just run into is the negation of a quantified statement. We have the following logical equivalences:
For all predicates $\varphi$
\begin{align*} \neg \forall x \varphi &= \exists x \neg \varphi \\ \neg \exists x \varphi &= \forall x \neg \varphi \\ \end{align*}We can consider this to be a generalized version of De Morgan's laws for quantifiers. This should be surprising! At first glance, we don't really have a reason to believe that $\forall$ and $\exists$ should be opposites - if anything, $\exists$ seems like a weaker equivalent of $\forall$.
This may seem strange, but it'll become intuitive when put in use. Taking our example $E(x)$ again, we may have a proposition $\forall x \: E(x)$, which says all numbers are even. This is, of course, false, because there exists a number that is not even. This is in contrast to, say $\forall x \: \neg E(x)$, a statement that is also false (why?) and therefore not the negation of the original proposition.
Let's take a moment to think about what this is saying.
Taking our example $\forall x (E(x) \rightarrow \neg E(x+1)$, we can think about what the negation of this statement might be. Of course, we can begin by slapping on a negation: $$\neg \forall x (E(x) \rightarrow \neg E(x+1)).$$ Is there anything more we can do to clarify what this statement means though? We can make use of the negation of quantified statements that were just mentioned: $$\exists x \neg(E(x) \rightarrow \neg E(x+1)).$$ From here, we can make use of the propositional equivalences we saw earlier. Applying the implication equivalence gets us $$\exists x (\neg (\neg E(x) \vee \neg E(x+1))).$$ Now we can use De Morgan's laws together and eliminate the resultant double negations to get $$\exists x (E(x) \wedge E(x+1)).$$ Now, this statement says that there exists an integer $x$ such that $x$ is even and $x+1$ is even.
The final statement we arrived at makes it much more clear what the negation of our original statement actually meant. Obviously, since it was so simple, we could have probably guessed what it meant without going through the formal process of applying inference rules. However, if you're faced with proving some proposition, it is helpful to try to methodically translate the statement to formal logic and consider what it is saying in this way to make it more clear what it is you need to prove.
That's all good and well, but you may be wondering how this information can directly be applied to proofs. You may especially be wondering about the point of filling in all these truth tables where we already know the values of $p$ and $q$: if we already know the truth value of $p$ and $q$, there's nothing to prove!
To see how to proceed, let's look back at our truth table for implications.
| $p$ | $q$ | $\quad$ | $p \rightarrow q$ |
|---|---|---|---|
| $T$ | $T$ | $T$ | |
| $T$ | $F$ | $F$ | |
| $F$ | $T$ | $T$ | |
| $F$ | $F$ | $T$ |
Side note: you may notice that $q$ also has value $T$ in the truth table when $p$ is false but $p \rightarrow q$ is true. However, the next row also states that when $p$ is false but $p \rightarrow q$ is true, $q$ can be false. What that tells us is that when $p$ is false but $p \rightarrow q$ is true, we can't actually conclude anything about $q$, which should make sense; the statement "if it rains, I'll bring an umbrella" doesn't actually tell us about what happens when it doesn't rain. As such, if we want to guarantee that $q$ is $T$, the only way to do so is to utilize the top row.
How, then, do we show that $p \rightarrow q$? Let's try an example. Say I am thinking of an integer $n$, and I tell you that it's even. You tell me that the square of my integer must also be even, but I don't believe you. In that case, you need to prove the following claim:
Breaking it down into propositions, if we define the preposition $E(x)$ to represent when $x$ is even, then what we need to prove is that $\forall n \: E(n) \rightarrow E(n^2)$. Here, to translate words into math, we say that \[E(n) = ``\text{There exists some integer }k\text{ such that }n=2k.\text{"}\]
Before we even get started, we run into a problem: the $\forall$ means we have to prove this true for every number. In the worst case scenario, this would mean that I have to take every possible value of $n$ and individually show that $E(n) \rightarrow E(n^2)$! I can show that $2^2 = 4$ is even, $4^2 = 16$ is even, $6^2 = 36$ is even, and so on, but of course this will take infinitely long.
To resolve this, we'll rely on the following logic: if I take an arbitrary $n$ that I don't know anything about, and prove that for that $n$, $E(n) \rightarrow E(n^2)$, then I can conclude that $\forall n \: E(n) \rightarrow E(n^2)$. This is known as the Principle of Universal Generalization. Now, I have to be careful here - I can't make any assumptions about $n$ beyond what I'm told. In this case, I only know that $n$ is an even integer. If I assume anything else about $n$, then I've lost my claim that this is true for all $n$, and can only argue it's true for the $n$ that follow my further assumption.
Consider the proposition: every student in this class has a CNetID. Of course, one way I could go about proving this is by checking every student individually. However, a faster way would be as follows. Take any student $s$ who has enrolled in the class. This can only be done with a CNetID. Thus $s$ has a CNetID.
Note that I only argued about one student $s$, but because it was a student chosen arbitrarily, I know that the proof applies to every student simultaneously, so I can conclude the "every" part of the proposition.
With the Principle of Universal Generalization in tow, we can finally start proving our proposition.
Select an even integer $x$. If $x$ is an even integer, there is some integer $k$ such that $x = 2k$. If $x = 2k$, then for that $k$, $x^2 = 4k^2$. If $x^2 = 4k^2$, then $x^2 = 2(2k^2)$. If $x^2 = 2(2k^2)$, then it can be written as the form $2k'$ for $k' = 2k^2$. If $x^2$ can be written in the form $2k'$, then $x^2$ is even. Thus, for any even integer $x$, $x^2$ is even.
To break down this proof, consider the following prepositions, keeping in mind that we have selected an arbitrary $x$:
With this notation, our proof becomes the following:
Select any $x$ where $p$. $p \rightarrow p_1$. $p_1 \rightarrow p_2$. $p_2 \rightarrow p_3$. $p_3 \rightarrow p_4$. $p_4 \rightarrow q$. Thus, for all $p$, $p \rightarrow q$.
Written this way, the proof is of course pretty illegible, but its structure becomes very clear. We're starting with $p$ and making a chain of implications that end in $q$. Of note, any of these individual implications should be very clearly true. For example, $p \rightarrow p_1$ and $p_4 \rightarrow q$ are just our definition of $E$, and the other implications are just one step of algebra, so it is easy to tell that they are true as well.
The really cautious among you may also notice that we assumed that implications are transitive - that because this chain exists, we can collapse it down to $p \rightarrow q$. The logic should of course intuitively make sense, but intuition isn't everything, so it can't hurt to check whether this is actually the case.
| $p$ | $q$ | $r$ | $p \to q$ | $q \to r$ | $(p \to q) \wedge (q \to r)$ | $p \to r$ | $((p \to q) \wedge (q \to r)) \to (p \to r)$ |
|---|---|---|---|---|---|---|---|
| $T$ | $T$ | $T$ | $T$ | $T$ | $T$ | $T$ | $T$ |
| $T$ | $T$ | $F$ | $T$ | $F$ | $F$ | $F$ | $T$ |
| $T$ | $F$ | $T$ | $F$ | $T$ | $F$ | $T$ | $T$ |
| $T$ | $F$ | $F$ | $F$ | $T$ | $F$ | $F$ | $T$ |
| $F$ | $T$ | $T$ | $T$ | $T$ | $T$ | $T$ | $T$ |
| $F$ | $T$ | $F$ | $T$ | $F$ | $F$ | $T$ | $T$ |
| $F$ | $F$ | $T$ | $T$ | $T$ | $T$ | $T$ | $T$ |
| $F$ | $F$ | $F$ | $T$ | $T$ | $T$ | $T$ | $T$ |
This style of proof is called a direct proof. In the rest of the unit, we'll talk more about how to construct these proofs, and what other types of proof exist.
The following problems from Rosen (8th edition) cover the topics from this lecture: