Math From Scratch: Creating a Proof Checker
Math, at its core, is a network of facts that are undoubtably known to be true. We start out with some simple assumptions called axioms and prove new facts to be true using logic and reasoning. As long as the proofs are valid, then under the founding assumptions, anything that can is proven is definitely true. But when you’re proving something in math, how do you actually know if you’re doing it right? What if you make a mistake? What if you gloss over proving something which is "obviously true" that turns out to be false? One way to be sure is to have a computer check your proof, which is what we’re going to do. But first, we’re going to have to build math from the ground up.



Proofs are essential to math. We start out with some facts we assume to be true called axioms. From those axioms, we can deduce other facts. If a fact is proven to be true given the axioms, it is called a theorem. Working in such a system, we can use theorems to prove other theorems. In this way, math builds on itself, using reasoning to accumulate more and more facts that follow from those initial assumptions. But this process relies on a few things to make sense: Your axioms and rules of reasoning must be "sensible", and you must apply them correctly in a proof. In this blog post, we will create a language for stating mathematical facts, a "sensible" set of rules for reasoning and, in the next post, a system that checks the correctness of proofs using those rules. We will be able to throw in whatever axioms we want and explore what facts we can prove from them.
For this post, you actually won’t need to know much math. It would help to be sort of familiar with proofs, logic, and sets, but we’re going to be building math from the ground up, so you won’t need much to start with.
In this post, we will lay the foundation for our proof checker by creating a language for mathematical statements and a set of rules for proving statements. In part 2, we will implement a proof checker based on our language and rules.
1 A Language for Facts
In order to state facts, we need a language to write those facts in. For example, let’s consider a statement written in English:
"All men are mortals. Socrates is a man. Therefore Socrates is mortal."
What exactly is the relationship between these sentences? What do "is" and "are" mean? You can look up the definitions for "is" and "are" in a dictionary, but those definitions are in terms of other imprecise english words. What we want are some welldefined connectives for building statements and relating objects. To this end, we will use the language of firstorder logic.
A formula is a statement written in firstorder logic. A formula \(P,Q\) is one of
A variable
\[x\]
A universal quantification
\[ \forall x (P) \]
An existential quantification
\[ \exists x (P) \]
An implication (ifthen)
\[ P \rightarrow Q \]
A conjunction (and)
\[ P \wedge Q \]
A disjunction (inclusive or)
\[ P \vee Q \]
A negation (not)
\[ \neg P \]
An operation. Like \(x \in y\) for set membership or \(x = y\) for equality.
A formula in firstorder logic represents a mathematical statement of a fact that may or may not be true. I’ll be using "formula" and "statement" somewhat interchangably, but remember that a statement is the idea of a fact, and a formula is a way of writing it down in our language. And not all formulas represent statements. Some formulas just represent objects that are being discussed in a statement.
Let’s go through each type of formula and think about what these statements mean:
A variable is a name that represents some object, like a set.
Universal quantification is used to make a statement about all things. \(\forall x (P)\) means "for all x, the statement P is true", where \(P\) is some statement about \(x\). Here, \(x\) is a variable in our language, and \(P\) is a metavariable representing some formula/statement in our language. This is a subtle distinction. When you see something like \(x,y,z,S\), that means a variable in our language, and when you see \(P,Q,R\), that means some formula, not literally the variable \(P\). So \(x\) is literally the formula \(x\), and \(P\) is a metavariable representing some formula in our language.
Existential quantification is used to say that there exists an object with some property. \(\exists x (P)\) means "there exists an x such that P is true", where \(P\) says something about \(x\).
Implication has a few uses. One is to make assumptions. \(P \rightarrow Q\) means that if you assume that \(P\) is true, then \(Q\) must be true. This sounds like how we use axioms. Another use is to state implications directly. In the Socrates example, we want to say "All men are mortals". To state that fact in our language, we can say \(\forall x (x \in men \rightarrow x \in mortals)\), which means that being in the set of men implies being in the set of mortals. In other words, being a man implies being a mortal. This pattern of an implication in a forall is very common. Often times, we don’t want to make a direct statement about all things, but rather all things with some property. \(\forall x (P \rightarrow Q)\) means "for all x where P is true, Q is true."
Conjunction (and) is used to state that two facts are both true. \(P \wedge Q\) is true when both \(P\) is true and \(Q\) is true.
Disjunction (inclusive or) is used to state that one fact or another fact are true, or both. \(P \vee Q\) is true when \(P\) is true, \(Q\) is true, or they’re both true.
A negation (not) is used to state that a fact is not true. \(\neg P\) is true when \(P\) is not true.
We mentioned set membership \(\in\), but we haven’t defined it. Operators like \(\in\) aren’t part of the fundamental language of firstorder logic, and the rules of logic are independent from the definitions of operators. We’ll see how they fit in later. I only include statements about sets in our examples so we can have something concrete to state facts about. Otherwise, it’s very abstract and hard to think about. For now, let’s just say that a set is a collection of objects and \(x \in y\) means that \(x\) is in the set \(y\).
Let’s look at some examples to get a feel for this language:
\[\forall x (\exists S (x \in S))\]
This means "For all x, there exists S such that x is an element of S". This is saying that for any object, there is a set containing that object. This is not saying that there is only one such set, just that there is at least one.
Here is another example:
\[\forall x (\forall S (x \in S \vee \neg (x \in S)))\]
This means "For all x, for all S, x is an element of S or x is not an element of S".
Here is our socrates example written in this language:
\[ (\forall x (x \in men \rightarrow x \in mortals) \wedge socrates \in men) \rightarrow socrates \in mortals \]
In English: "If, for all things x, x being a member of the set of men implies that x is a member of the set of mortals and socrates is in the set of men, then socrates is in the set of mortals." Simplifying a little more, "If being in the set of men implies being in the set of mortals and socrates is in the set of men, then socrates is in the set of mortals". We state "socrates is mortal" with the formula \(socrates \in mortals\), meaning "socrates is in the set of mortals".
Let’s think about what the variables in our statements really mean. In examples using set membership, variables represented sets and elements of sets. But we can also make statements like this:
\[\forall x (x \rightarrow x)\]
This means that "for all x, if x is true, then x is true". This is a statement about statements, rather than a statement about sets. And it feels like it should be true, so it’s not nonsense. In fact, this is a completely valid statement. But if variables can represent statements and also objects like sets, then we can write nonsense like
\[ \forall x (\forall S (x \in (\neg S))) \]
This is treating \(S\) as a statement and as a set, which is nonsense. Saying that an object is an element of a statement is meaningless.
It’s actually fine that we can write nonsense like this. We just have to make sure that our axioms and rules for reasoning don’t allow us to prove nonsensical statements like that to be true. Or we can create a system in which statements like this actually have some meaning, which is also fine. We just have to keep in mind that variables can represent statements as well as other mathematical objects like sets.
2 Rules for Reasoning
We now have a language for stating mathematical facts. Now, we need a set of rules for proving that statements are true. These are our rules for reasoning.
In order to prove anything interesting, we need to start out with some assumptions. Throughout a proof, we’ll have some collection of assumptions which we’ll use to prove some statements. We will represent this collection of assumptions with a context, often denoted with the greek letter \(\Gamma\), pronounced "gamma". A context is a list of formulas which are assumed to be true statements. For example, in the socrates example, we assumed that all men are mortals and that socrates is a man. We could represent these assumptions as a context like this:
\[ \Gamma = \forall x (x \in men \rightarrow x \in mortals), socrates \in men \]
In this case, \(\Gamma\) is a list of two assumptions. If we want to make a statement under those assumptions, we’d write it like this:
\[ \forall x (x \in men \rightarrow x \in mortals), socrates \in men \vdash socrates \in mortals \]
The turnstile symbol \(\vdash\) separates the assumptions from the statement you’re making. In general,
\[ \Gamma \vdash P \]
means under the assumptions in the context \(\Gamma\), \(P\) is true. This is called a judgement. Rather than proving statements directly, we will be proving judgements. For a theorem, \(\Gamma\) would contain your axioms and \(P\) would be the statement of your theorem. Proving a theorem is proving your statement to be true assuming that the axioms are true.
Let’s think back to our original statement of the socrates example:
\[ (\forall x (x \in men \rightarrow x \in mortals) \wedge socrates \in men) \rightarrow socrates \in mortals \]
One of the uses of implication is to encode the idea of an assumption. Isn’t this the same as our judgement? Yes! In general, the judgement
\[ P,Q,\ldots \vdash R \]
is equivalent to the statement
\[ P \wedge Q \wedge \ldots \rightarrow R \]
We will represent this fact with two rules. One for proving an implication:
\[ \frac{ \Gamma, P \vdash Q }{ \Gamma \vdash P \rightarrow Q } ({\rightarrow} R) \]
and one for using a conjunction:
\[ \frac{ \Gamma, P, Q \vdash R }{ \Gamma, P \wedge Q \vdash R } ({\wedge} L) \]
Let’s start out by talking about this notation for rules. We have two judgements separated by a line called an inference line, and we have the name of the rule to the right of the whole thing, like \(({\rightarrow} R)\). The meaning of a rule written this way is, in order to prove the bottom judgement, you must prove the top judgement. I find it easiest to read these rules starting at the bottom and then the top. For our implication rule \(({\rightarrow} R)\), we are saying that in order to prove that an implication \(P \rightarrow Q\) is true using some assumptions \(\Gamma\), it suffices to prove that \(Q\) is true if we add \(P\) to our list of assumptions. This is how we use implication to represent assumptions. It is literally baked in to our rules for reasoning.
Now let’s talk about that second rule, \(({\wedge} L)\). This is saying that if we assume that a conjunction \(P \wedge Q\) is true, we can assume that both statements are true and add both statements to our list of assumptions.
A proof in this system will consist of a sequence of applications of rules until we reach a judgement that requires no subproofs to prove. We’ll see how that works soon. This is all pretty abstract, so let’s bring it back to our socrates example and sketch out a proof of that statement:
We start out with this judgement:
\[ \vdash (\forall x (x \in men \rightarrow x \in mortals) \wedge socrates \in men) \rightarrow socrates \in mortals \]
Notice that there is nothing to the left of the turnstile. This means we are starting out with no assumptions. Now, to start proving this statement, we can use our rules:
\[ \displaystyle \frac{ \displaystyle \frac{ \displaystyle \frac{ \displaystyle \cdots }{ \displaystyle \forall x (x \in men \rightarrow x \in mortals), socrates \in men \vdash socrates \in mortals } (?) } { \displaystyle \forall x (x \in men \rightarrow x \in mortals) \wedge socrates \in men \vdash socrates \in mortals } ({\wedge} L) }{ \displaystyle \vdash (\forall x (x \in men \rightarrow x \in mortals) \wedge socrates \in men) \rightarrow socrates \in mortals } ({\rightarrow} R) \]
Remember, we read bottom up and the proof grows upward as we progress. We start out with no assumptions. We must prove an implication, so we use \(({\rightarrow} R)\). Then, we have a conjunction in our assumptions, so we use \(({\wedge} L)\) to add both statements to our assumptions. We don’t know what to do next, so we’ll leave it for now.
I chose to name these rules \(({\rightarrow} R)\) and \(({\wedge} L)\) for a reason. You use \(({\rightarrow} R)\) to prove that an implication is true, and you use \(({\wedge} L)\) to use a conjunction in your assumptions. In fact, for all of our different types of formulas, there is an \(L\) rule for using it as an assumption and an \(R\) rule for proving that the statement is true. To make some more progress in our Socrates proof, let’s define how we can use a universal quantification.
Notice that we "lost" the conjunction assumption when we used \(({\wedge}L)\). In this case, that’s fine. But for other \(L\) rules, we might want to use an assumption more than once. Later on, we’ll introduce a rule for duplicating assumptions.
If we assume that a universal quantification \(\forall x (P)\) is true, that means the statement \(P\) is true for any thing \(x\). In our example, we assume
\[ \forall x (x \in men \rightarrow x \in mortals) \]
If we assume \(\forall x (x \in men \rightarrow x \in mortals)\), then we are assuming that this is true for all things. So this should also be true for socrates! In other words, we should be able to assume that
\[ socrates \in men \rightarrow socrates \in mortals \]
What we just did is substitute \(socrates\) for \(x\) in the body of the quantification. In general, if we assume a universal quantification \(\forall x (P)\) to be true, we should be able to substitute anything for \(x\) in \(P\) and assume that to be true. This is the \(({\forall} L)\) rule:
\[ \frac{ \Gamma, P[t \mathbin{/} x] \vdash Q } { \Gamma, \forall x (P) \vdash Q } ({\forall} L) \]
where \(P[t \mathbin{/} x]\) denotes substituting the formula \(t\) in place of the variable \(x\) in \(P\). \(t\) does not necessarily represent a statement, so we refer to it as a "term". Here, \(t\) can be any formula, not just a variable.
This is sometimes referred to as "instantiating" the universal quantification with a particular term \(t\).
Cool! Let’s use this in our Socrates proof:
\[ \displaystyle \frac{ \displaystyle \frac{ \displaystyle \frac{ \displaystyle \frac{ \displaystyle \cdots }{ \displaystyle socrates \in men \rightarrow socrates \in mortals, socrates \in men \vdash socrates \in mortals } (?) }{ \displaystyle \forall x (x \in men \rightarrow x \in mortals), socrates \in men \vdash socrates \in mortals } ({\forall} L) }{ \displaystyle \forall x (x \in men \rightarrow x \in mortals) \wedge socrates \in men \vdash socrates \in mortals } ({\wedge} L) }{ \displaystyle \vdash (\forall x (x \in men \rightarrow x \in mortals) \wedge socrates \in men) \rightarrow socrates \in mortals } ({\rightarrow} R) \]
Our \(({\forall} L)\) rule is written as if the universal quantification has to be the last element of the context, but order of assumptions doesn’t matter, so it can be anywhere in the context for the rule to be applicable. Later on, we’ll justify this be adding a rule that allows us to reorder the asusmptions of the context.
We’re one step closer! Now we have an implication in our context of assumptions. How do we use that?
Remember, the direct interpretation of the implication \(P \rightarrow Q\) being true is that \(P\) being true implies that \(Q\) is true. In other words, if we assume that \(P \rightarrow Q\) is true and we prove that \(P\) is true, then we should be able to assume that \(Q\) is true. This is the \(({\rightarrow L})\) rule:
\[ \frac{ \Gamma \vdash P \qquad \Gamma, Q \vdash R }{ \Gamma, P \rightarrow Q \vdash R } ({\rightarrow} L) \]
This means that when we assume an implication \(P \rightarrow Q \) and we’re trying to prove that some statement \(R\) is true, if we can prove the antecedent \(P\) is true, then we can assume the consequent \(Q\) is true and use it to prove \(R\). This is an example of a rule which requires two subproofs instead of just one. We have to prove \(P\) before we can assume \(Q\) to prove \(R\).
Let’s use it in our proof:
\[ \displaystyle \frac{ \displaystyle \frac{ \displaystyle \frac{ \displaystyle \frac{ \displaystyle \frac{\cdots}{socrates \in men \vdash socrates \in men} (?) \qquad \frac{\cdots}{socrates \in mortals, socrates \in men \vdash socrates \in mortals} (?) }{ \displaystyle socrates \in men \rightarrow socrates \in mortals, socrates \in men \vdash socrates \in mortals } ({\rightarrow} L) }{ \displaystyle \forall x (x \in men \rightarrow x \in mortals), socrates \in men \vdash socrates \in mortals } ({\forall} L) }{ \displaystyle \forall x (x \in men \rightarrow x \in mortals) \wedge socrates \in men \vdash socrates \in mortals } ({\wedge} L) }{ \displaystyle \vdash (\forall x (x \in men \rightarrow x \in mortals) \wedge socrates \in men) \rightarrow socrates \in mortals } ({\rightarrow} R) \]
We’re almost done!
For our remaining subproofs, the statement we’re trying to prove is actually in the context of assumptions! If we assume that a statement is true, we don’t need to do any more work to prove that it is true. This is the \((I)\) rule.
\[ \frac{}{\Gamma,P \vdash P} (I) \]
Notice that there are no subproofs. In fact, all valid proofs will end with rules like this one, which have no subproofs. In this way, a proof is a tree of inferences, where each node is a judgement and an application of a rule, and its children are the subproofs required by the rule. The leaves of this tree are applications of rules like \((I)\) which require no subproofs. We’ll come back to this fact when we implement our proof checker. For now, let’s just realize that proofs are treelike.
Let’s finish up our proof:
\[ \displaystyle \frac{ \displaystyle \frac{ \displaystyle \frac{ \displaystyle \frac{ \displaystyle \frac{}{socrates \in men \vdash socrates \in men} (I) \qquad \frac{}{socrates \in mortals, socrates \in men \vdash socrates \in mortals} (I) }{ \displaystyle socrates \in men \rightarrow socrates \in mortals, socrates \in men \vdash socrates \in mortals } ({\rightarrow} L) }{ \displaystyle \forall x (x \in men \rightarrow x \in mortals), socrates \in men \vdash socrates \in mortals } ({\forall} L) }{ \displaystyle \forall x (x \in men \rightarrow x \in mortals) \wedge socrates \in men \vdash socrates \in mortals } ({\wedge} L) }{ \displaystyle \vdash (\forall x (x \in men \rightarrow x \in mortals) \wedge socrates \in men) \rightarrow socrates \in mortals } ({\rightarrow} R) \]
We did it! Now we know for sure that Socrates is mortal! That is, as long as our assumptions are true and our rules for reasoning make sense.
We started out with an imprecise english statement of a fact. Then, we translated the statement into a precise mathematical language. Finally, we have proven that the statement is true using a set of simple rules that make intuitive sense.
Now, let’s write down the rules for all the types of formulas:
We already figured out \(({\forall} L)\):
\[ \frac{ \Gamma, P[t \mathbin{/} x] \vdash Q } { \Gamma, \forall x (P) \vdash Q } ({\forall} L) \]
What about \(({\forall} R)\)? How do we prove that something is true about all things? In order to do that, we can instantiate the forall with some new variable that has no current meaning and no assumptions about it. Here’s what this looks like:
\[ \frac{ \Gamma \vdash P[y \mathbin{/} x] }{ \Gamma \vdash \forall x (P) } ({\forall} R) \]
We must add the restriction that \(y\) must not be mentioned anywhere in our context of assumptions \(\Gamma\) or our statement \(\forall x (P)\). More precisely, \(y\) must not occur free in \(\Gamma\) or \(\forall x (P)\). We’ll define what that means later. This restriction is necessary because if we are proving that \(P\) is true for all \(x\), we must be able to prove \(P[y \mathbin{/} x]\) without knowing or assuming anything about \(y\). This makes sense because if we prove that it is true for \(y\) without knowing anything about \(y\), then \(y\) could’ve been anything and we could’ve used the exact same proof.
For example,
\[ \displaystyle \frac{ \displaystyle \frac{ \displaystyle \frac{ }{ \displaystyle y \vdash y } (I) }{ \displaystyle \vdash y \rightarrow y } ({\rightarrow} R) }{ \displaystyle \vdash \forall x (x \rightarrow x) } ({\forall} R) \]
If \(y\) was some other statement, even nonsense, we could’ve used the exact same proof structure:
\[ \displaystyle \frac{ \displaystyle \frac{ }{ \displaystyle (x \in \neg x) \vdash (x \in \neg x) } (I) }{ \displaystyle \vdash (x \in \neg x) \rightarrow (x \in \neg x) } ({\rightarrow} R) \displaystyle \]
Now let’s do \({\exists} L\). If we assume that \(\exists x (P)\) is true, then we know that there exists at least one thing \(x\) such that \(P\) is true. How can we use this assumption? We can instantiate the quantifier! We can introduce some new variable \(y\) representing the thing that exists, and add an assumption \(P[y \mathbin{/} x]\) stating that the statement \(P\) is true for \(y\). Similar to \(({\forall} R)\), \(y\) must not occur free in the context \(\Gamma\) or \(\exists x (P)\). All we are allowed to assume about \(y\) is that \(P[y \mathbin{/} x]\) is true. That’s all we know about it.
\[ \frac{ \Gamma, P[y \mathbin{/} x] \vdash Q }{ \Gamma, \exists x (P) \vdash q } ({\exists} L) \]
This rule is dual to \(({\forall} R)\). It’s the same thing, except everything is happening to the left of the turnstile instead of the right. Proving a \(\forall\) is dual to using an \(\exists\). Pretty cool!
Unsurprisingly, the rule \(({\exists} R)\) is dual to \(({\forall} L)\):
\[ \frac{ \Gamma \vdash P[t \mathbin{/} x] }{ \Gamma \vdash \exists x (P) } ({\exists} R) \]
In order to prove an existential \(\exists x (P)\), you must substitute some term \(t\) for \(x\) in \(P\) and prove that \(t\) makes the statement true. In other words, to prove that something exists with a property, you give an example and prove that it has the property! And for the duality, proving an \(\exists\) is dual to using a \(\forall\).
We figured out \({\wedge} L\). What about \(({\wedge} R)\)? How do we prove \(P \wedge Q\)? We prove \(P\) and we prove \(Q\)!
\[ \frac{ \Gamma \vdash P \qquad \Gamma \vdash Q }{ \Gamma \vdash P \wedge Q } ({\wedge} R) \]
Makes sense. Unsurprisingly, the disjuncion rules are dual to these conjunction rules:
\[ \frac{ \Gamma,P \vdash R \qquad \Gamma,Q \vdash R }{ \Gamma,P \vee Q \vdash R } ({\vee} L) \]
This one is a little confusing on first glance. Let’s think about what this means: We assume \(P \vee Q\) is true. So we assume that either \(P\) is true, \(Q\) is true, or they’re both true. If we want to use this assumption, we need to handle each possible case (the case where they’re both true is implicitly handled). Thus, we need a separate proof for each case.
\(({\vee} R)\) actually has two rules:
\[ \frac{ \Gamma \vdash P }{ \Gamma \vdash P \vee Q } ({\vee} R1) \]
\[ \frac{ \Gamma \vdash Q }{ \Gamma \vdash P \vee Q } ({\vee} R2) \]
Since a disjunction is true when either statement is true, proving a disjunction only requires proving one of the statements. In a proof, you can choose whichever statement is easier to prove.
Why isn’t this dual to \(({\wedge} L)\)? It could have been. We could have written that as two rules as well:
\[ \frac{ \Gamma, P, \vdash R }{ \Gamma, P \wedge Q \vdash R } ({\wedge} L1) \]
\[ \frac{ \Gamma, Q, \vdash R }{ \Gamma, P \wedge Q \vdash R } ({\wedge} L2) \]
But, with another rule we’ll introduce later which allows us to duplicate assumptions, we’ll see that \(({\wedge} L)\) is equivalent to applying that duplication rule, \(({\wedge} L1)\), and then \(({\wedge} L2)\). So we might as well always do both.
We already have both rules for implication.
All that’s left is negation, which is a little weird. Negation says that a statement is false. How can we use that as an assumption? If we assume that a statement is false and show that this assumption leads to that statement being true, we have reached a contradiction. The statement can’t be both true and false. If we can reach a contradiction, all bets are off.
\[ \frac{ \Gamma \vdash P }{ \Gamma, \neg P \vdash Q } ({\neg} L) \]
If we assume that \(P\) is false and prove that it is true, we have a contradiction. We can use this contradiction to prove anything. This can be seen by the fact that we prove \(Q\), which could be anything, by proving \(P\).
And how do we prove a negation \(\neg P\)?
\[ \frac{ \Gamma, P \vdash \forall x (x) }{ \Gamma \vdash \neg P } ({\neg} R) \]
We assume that \(P\) is true and prove that this leads to a contradiction. This means it must be false. We choose \(\forall x (x)\) to represent "false" or "contradiction" because it’s the simplest statement that is impossible to prove unless you reach a contradiction like \(x \wedge \neg x\). This rule is the essence of a proof by contradiction.
Now, we have rules for every type of formula. One for using it as an assumption and one for proving it (actually, two for proving \(\vee\)). We also have a special rule, \((I)\), for when the statement we want to prove is already assumed to be true.
I also mentioned some rules for duplicating and reordering assumptions. These are called structural rules. Here they are:
\[ \frac{ \Gamma,P,P \vdash Q }{ \Gamma,P \vdash Q } (CL) \]
This allows us to duplicate assumptions so we don’t have to "lose" them when we use them. All of our \(L\) rules replace the used assumption with something else. But the more assumptions we have, the easier it is to prove things. So we will implicitly invoke this rule when we invoke \(L\) rules to avoid losing assumptions.
\[ \frac{ \Gamma_1,Q,P,\Gamma_2 \vdash R }{ \Gamma_1,P,Q,\Gamma_2 \vdash R } (PL) \]
This allows us to permute, or reorder our assumptions. In other words, the order of assumptions does not matter.
These two rules allow us to treat the context of assumptions as a set where we pretty much just keep adding assumptions as we go up the inference tree.
We can also forget assumptions if we don’t need them:
\[ \frac{ \Gamma \vdash Q }{ \Gamma,P \vdash Q } (WL) \]
This is valid because if we can prove that \(Q\) is true without assuming \(P\), then that is also a valid proof if you assume \(P\).
Another rule we can add for convenience is the \((Cut)\) rule:
\[ \frac{ \Gamma \vdash Q \qquad \Gamma,Q \vdash P }{ \Gamma \vdash P } (Cut) \]
This means if we prove some auxiliary statement \(Q\), often called a lemma, we can use it as an assumption to prove our original statement \(P\). This rule is technically not necessary. It can be shown that any proof that uses it can be translated to a proof that doesn’t use it. But is often convenient to have and can save us from duplicating proofs. This rule also allows us to build up our knowledge. For example, if we think of \(Q\) as a theorem, this means if we prove a theorem, we can assume that the theorem is true to prove new theorems.
Let’s think a little more about \(\forall x (x)\). This is saying that all statements are true, which is obviously false. If we can prove that this statement is true, and thus use it as an assumption with \((Cut)\), we can use \(({\forall}L)\) to prove anything. This would essentially be proving that false is true, which is a contradiction. This is an example of why a contradiction can be used to prove anything, which hopefully makes \(({\neg}L)\) seem less weird.
For those who want to learn more about this kind of system of rules, look up "Sequent Calculus".
2.1 Tying up Loose Ends
Now, let’s tie up some loose ends. First, let’s talk about free variables and substitution.
In the formula \(\forall x (P)\), we say that \(x\) is bound in \(P\). In other words, \(x\) is in scope for the formula \(P\). This means that when \(P\) references the variable \(x\), it is specifically referring to the universally quantified variable \(x\). For example, in the formula \(\forall x (x \rightarrow x)\), the \(x\) in the implication is the universally quantified \(x\). In contrast, if we consider the formula \(x \wedge y\) in isolation, \(x\) and \(y\) have no meaning. They are free variables. They are not bound. We say that \(x\) occurs free in \(x \wedge y\) and \(x\) occurs bound in \(\forall x (x \rightarrow x)\). In other words, a variable occurs free in an expression when it is referenced in a place where it is not bound by anything like a quantification.
Let’s define this for each type of formula:
Let’s say we’re asking whether the variable \(x\) occurs free in some formula.
For a variable formula, \(x\) occurs free in the formula if the formula is \(x\). For example, \(x\) occurs free in the formula \(x\), but \(x\) does not occur free in the formula \(y\).
For a quantification, if the name of the quantified variable is \(x\), then \(x\) does not occur free. For example, \(x\) does not occur free in \(\forall x (x \rightarrow x)\) because its references are bound by the quantifier. In contrast, \(x\) does occur free in \(\forall y (x \rightarrow x)\) because the quantifier does not bind \(x\). So, in general, \(x\) occurs free in a quantifier if the quantified variable is not \(x\) and \(x\) occurs free in the body formula.
For a connective or an operator like \(P \rightarrow Q\), \(x\) occurs free in that formula if \(x\) occurs free in any of the subformulas.
Here are some more examples:
The variables \(x\) and \(y\) both occur free in \(x \wedge y\).
The variable \(z\) does not occur free in \(x \wedge y\).
The variable \(x\) does not occur free in \(\forall x (x \wedge y)\), but the variable \(y\) does.
The variable \(x\) occurs free in \(x \wedge \forall x (x)\). Although it is bound in the forall, it is also referenced outside of the forall, where it is free.
A variable occurs free in a context if it occurs free in any formula in the context.
Hopefully, the restrictions on \(({\forall} R)\) and \(({\exists} L)\) make sense now. If \(y\) occurred free in the context, we’d know something about \(y\).
Now, let’s talk about substitution. Recall that the notation for substitution is \(P[t \mathbin{/} x]\), meaning replace \(x\) with the term \(t\) in the formula \(P\). For example, \((x \wedge y)[z \rightarrow z \mathbin{/} x]\) would be \((z \rightarrow z) \wedge y\). Let’s define substitution:
For a variable formula, \(x[t \mathbin{/} x]\) becomes \(t\). But if the variable is different, \(y[t \mathbin{/} x]\) is just \(y\).
For a quantification, \((\forall y (P))[t \mathbin{/} x]\) becomes \(\forall y (P[t \mathbin{/} x])\). We just substitute in the body. But if the quantified variable is \(x\), we don’t substitute. So \((\forall x (P))[t \mathbin{/} x]\) stays \(\forall x (P)\). This is because we only want to substitute free variables. Since \(x\) takes on a new meaning in the body of the quantifier, it shouldn’t get substituted. Another problem is that the quantified variable could occur free in \(t\). For example, in the subtitution \((\forall y (y \wedge x))[y \mathbin{/} x]\), we get \(\forall y (y \wedge y)\). We substituted a free variable \(y\) for \(x\), but it ended up bound after the subtitution. This is a problem because \(y\) could mean something outside of this quantification, but when it gets substituted in, its meaning gets changed to the universally quantified variable. This problem is known as name capture. For now, we will just say that this substitution is illegal.
For a connective or operator like \(P \rightarrow Q\), we just substitute in each subformula.
Another thing we briefly mentioned is equivalence. What does it mean for two statements to be equivalent? There are a few ways to interpret this. We can define equivalence to mean that two formulas must be identical. But this is unnecessarily restrictive. Under this definition, \(x \wedge y\) would not be equivalent to \(y \wedge x\) even though those two statements feel like they should be equivalent. Another way of defining equivalence is that \(P\) and \(Q\) are equivalent if \(P\) is true when \(Q\) is true and vice versa. Symbolically, \((P \rightarrow Q) \wedge (Q \rightarrow P)\). This also means that if one statement is false, the other must be false. Our equivalence statement is true when both statements are true or when both statements are false. This is called a bidirectional implication, written \(P \leftrightarrow Q\), and this is our notion of equivalent statements. We can prove that statements like \(x \wedge y\) and \(y \wedge x\) are equivalent by proving that bidirectional implication. People often read \(\leftrightarrow\) as "if and only if".
This notion of equivalence is useful, but it only makes sense for statements because implication and conjunction only make sense for statements. What if we want to say that two sets are the same? Or that two numbers are the same? We need a notion of equality to state that two objects are the same. How should we define equality?
First let’s define how we can write down an equality. We will write an equality like \(t_1 = t_2\), where \(t_1\) and \(t_2\) are "terms". A term is just a formula, but we use the word "term" when the formula might represent some object like a set or a number, not necessarily a statement.
Ok, now that we have a way to state an equality, how should we define it? Let’s start by listing some properties we want equality to have: Equality should be reflexive. So \(x = x\). Equality should be symmetric, so \(x = y \rightarrow y = x\). Equality should also be transitive, so \(x = y \wedge y = z \rightarrow x = z\). Another useful property to have is the subsitution property of equality. For example, if we have two equations, \(x = y + 1\) and \(x + y = 2\), we should be able to substitute the first equation into the second and get \(y + 1 + y = 2\). Now that we have some general idea of how we want our equality operator to behave, let’s write some rules:
A good first step is to make \(L\) and \(R\) rules, just like our logical connectives. Let’s start with \(({=}R)\). How do we prove that two things are equal? In the simplest case, it should always be true that something is equal to itself. That’s our reflexive property. As a rule, we write
\[ \frac{}{\Gamma \vdash t = t} ({=}R) \]
Like \((I)\), this requires no subproofs. Before we think about other ways we might prove that two things are equal, let’s think about how we can use an assumed equality. If we assume that \(x = y\), we can use the substitution property of equality to substitute \(x\) with \(y\) or vice versa. Anywhere we see an \(x\), we can replace it with a \(y\). In a proof, we’d run this substitution on the statement we’re trying to prove. Here is the rule:
\[ \frac{\Gamma \vdash P[t \mathbin{/} x]}{\Gamma,x=t \vdash P} ({=}L) \]
What about symmetry and transitivity? They actually follow from these two rules! Let’s prove that this is true, starting with symmetry:
\[ \displaystyle \frac{ \displaystyle \frac{}{\vdash y=y} ({=}R) }{ \displaystyle x=y \vdash y=x } ({=}L) \]
We substituted \(y\) for \(x\) in the equation we’re trying to prove. Now for transitivity:
\[ \displaystyle \frac{ \displaystyle \frac{ }{ y=z \vdash y=z } (I) }{ \displaystyle x=y,y=z \vdash x=z } ({=}L) \]
Cool! But there is a slight problem. Since substitution is only defined for replacing a variable, if we want to use an equation with nonvariable terms on both sides like \(x + y = y + x\), we wouldn’t be able to substitute these two terms for each other. To fix this, let’s extend our definition of substitution to be able to replace terms as well as variables.
We will add a case: \(t_1[t_2 \mathbin{/} t_1]\) becomes \(t_2\). If the whole formula is identical to the term we’re replacing, we just replace the whole formula. For example, \((x \rightarrow y)[a \wedge b \mathbin{/} x \rightarrow y]\) becomes \(a \wedge b\).
The variable case and the simple connectives cases stay the same (although the variable case is no longer necessary). For quantifiers, the rule \(\forall x (P)[t \mathbin{/} x]\) remains unchanged, but \(\forall y (P)[t \mathbin{/} x]\) becomes \(\forall y (P[t \mathbin{/} x])\). We needed to add that case where the variable being replace is the same as the quantified variable because we only wanted to substitute free occurrences of the variable, and the quantifier binds the variable, giving it a new meaning. To generalize this to term subsitution, \((\forall x (P))[t_2 \mathbin{/} t_1]\) will remain unchanged if \(x\) occurs free in \(t_1\), otherwise it will become \(\forall x (P[t_2 \mathbin{/} t_1])\). For example, \((\forall x (x \wedge (x \rightarrow y)))[z \mathbin{/} x \rightarrow y]\) remains \(\forall x (x \wedge (x \rightarrow y))\) because \(x\) occurs free in \(x \rightarrow y\). And again, to avoid name capture, the substitution is illegal if \(x\) occurs free in \(t_2\).
Now that we have extended our definition of substitution, we can write a more general \(({=}L)\) rule:
\[ \frac{\Gamma \vdash P[t_2 \mathbin{/} t_1]}{\Gamma,t_1=t_2 \vdash P} ({=}L) \]
We just extended our language to include equality and all we needed to do was add a rule for how to prove an equality and a rule for how to use an equality. This is a common pattern.
We now have a language for stating mathematical facts, a set of rules for proving statements in that language, and an idea of how to extend it as needed to incorporate new kinds of statements. Now, let’s explore how to apply this system to do some math!
3 Theories
3.1 Set Theory
Let’s talk about sets. What is a set? Intuitively, we think of a set as an unordered collection of things. Something is either in a set or not in a set. We will write set membership like \(x \in S\), which means \(x\) is an element of the set \(S\). This is a new type of formula in our language. Last time we added something new, equality, we added rules for how to prove it and how to use it as an assumption. For set membership, we won’t be doing that. You can’t really use the fact that an object is in a set on its own like you could with an equality. There is also no inherent way to prove set membership like there is for equality. Instead, we will write some axioms that describe the nature of set membership. This is how we will define what a set is. We will be using some of the axioms of ZermeloFraenkel set theory.
First up, we have the axiom of extensionality:
\[ \forall x (\forall y ([\forall z (z \in x \leftrightarrow z \in y)] \rightarrow x = y)) \]
That’s pretty dense. Let’s translate to English: For any two objects \(x\) and \(y\), if, for all objects \(z\), \(z \in x\) if and only if \(z \in y\), then \(x\) and \(y\) are the same set. In other words, two sets are equal when they have the same elements.
Next, we have the axiom of empty set:
\[ \exists \emptyset (\forall x (\neg (x \in \emptyset))) \]
This states that there exists a set with no elements.
Next, we have the axiom of pairing:
\[ \forall x (\forall y (\exists z (\forall e (e \in z \leftrightarrow e = x \vee e = y)))) \]
This is saying that, for any two things \(x\) and \(y\), there exists a set \(z\) containing exactly those two objects.
Next, we have the axiom of union:
\[ \forall F (\exists A (\forall Y (\forall x ((x \in Y \wedge Y \in F) \leftrightarrow x \in A)))) \]
Here, \(F\) is some set of sets, \(Y\) is an element of \(F\), \(x\) is an element of \(Y\), and \(A\) is the union of sets that are elements of \(F\). All this is saying is that for all sets of sets \(F\), there exists some set \(A\) that is the union of all sets that are elements of \(F\). For example, if we had \(F = \{\{a,b,c\}, \{b,c,d\}, \{a\}\}\), we’d have \(A = \{a,b,c,d\}\). The property that defines \(A\) is that elements of \(A\) are elements of elements of \(F\).
Next, we have the axiom of power set:
\[ \forall x (\exists y (\forall z ([\forall e (e \in z \rightarrow e \in x)] \leftrightarrow z \in y))) \]
The power set of a set \(x\) is the set of all subsets of \(x\). \(z\) is a subset of \(x\) if and only if every element of \(x\) is an element of \(z\). That’s what \(\forall e (e \in z \rightarrow e \in x)\) is saying. For example, the power set of \(\{1,2,3\}\) is \(\{\{\}, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}\}\). Notice that the empty set \(\{\}\) and the set itself \(\{1,2,3\}\) are subsets of \(\{1,2,3\}\). This axiom is saying that for all sets \(x\), the power set of \(x\), \(y\), exists.
Next, we have the axiom schema of specification:
Let \(P\) be some formula with free variables \(x,z,w_1,w_2,\ldots,w_n\) where \(y\) does not occur free in \(P\).
\[ \forall z (\forall w_1 (\forall w_2 (\ldots \forall w_n (\exists y (\forall x (x \in y \leftrightarrow (x \in z \wedge P))))))) \]
This is hard to read, but the idea is actually quite simple. If we have some set \(z\), there exists a set \(y\) which is a subset of \(z\) that contains elements of \(z\) that have some property \(P\). We often see this written like \(y = \{x \in z : P\}\). In other words, we can construct a set \(y\) from a set \(z\) by "filtering" its elements. You might be wondering why we can’t just construct a set \(\{x : P\}\) of all objects with some property. If we allowed that, we could write \(\{x : \neg (x \in x)\}\). This is the set of all sets that do not contain themselves. This is Russel’s paradox and it leads to a contradiction. To avoid this, we restrict ourselves to constructing subsets of existing sets.
This is called an axiom schema because there is an axiom for every formula \(P\) where \(y\) does not occur free.
There are more axioms in ZermeloFraenkel set theory, but we’ll stick to these for now.
These axioms describe what equality means for sets and states what sets can exist. We can express ideas like subsets, power sets, intersections, and unions using logic and our membership operator.
How can we use these axioms to prove things about sets? For the axioms, we can just put them in our context of assumptions. For an axiom schema like the axiom schema of specification, we can use a rule:
\[ \frac{ \Gamma,\exists y (\forall x (x \in y \leftrightarrow (x \in z \wedge P))) \vdash Q }{ \Gamma \vdash Q } (Spec) \]
With the restriction that \(y\) does not occur free in \(P\)
This allows us to add the axiom (with the foralls of \(z\) and the free variables already instantiated) for some formula \(p\) to our context at any time.
As an example, let’s prove the existence of binary set intersections. That is, given two sets, there exists a set whose elements are in both sets. We’ll use the axiom schema of specification.
\[ \displaystyle \frac{ \displaystyle \frac{ \displaystyle \frac{ \displaystyle \frac{ \displaystyle \frac{ \displaystyle \frac{ \displaystyle \frac{ \displaystyle \frac{}{ \displaystyle d \in c \leftrightarrow (w \in a \wedge w \in b) \vdash d \in c \leftrightarrow (w \in a \wedge w \in b) } (I) }{ \displaystyle \forall x (x \in c \leftrightarrow (x \in a \wedge x \in b)) \vdash d \in c \leftrightarrow (w \in a \wedge w \in b) } ({\forall}L) }{ \displaystyle \forall x (x \in c \leftrightarrow (x \in a \wedge x \in b)) \vdash \forall w (w \in c \leftrightarrow (w \in a \wedge w \in b)) } ({\forall}R) }{ \displaystyle \forall x (x \in c \leftrightarrow (x \in a \wedge x \in b)) \vdash \exists z (\forall w (w \in z \leftrightarrow (w \in a \wedge w \in b))) } ({\exists}R) }{ \displaystyle \exists y (\forall x (x \in y \leftrightarrow (x \in a \wedge x \in b))) \vdash \exists z (\forall w (w \in z \leftrightarrow (w \in a \wedge w \in b))) } ({\exists}L) }{ \displaystyle \vdash \exists z (\forall w (w \in z \leftrightarrow (w \in a \wedge w \in b))) } (Spec) }{ \displaystyle \vdash \forall y (\exists z (\forall w (w \in z \leftrightarrow (w \in a \wedge w \in y)))) } ({\forall}R) }{ \displaystyle \vdash \forall x (\forall y (\exists z (\forall w (w \in z \leftrightarrow (w \in x \wedge w \in y))))) } ({\forall}R) \]
We used specification on the set \(a\) and our property \(P\) was \(x \in b\).
You might be wondering where symbols like \(\subseteq\) and curly brace notation \(\{\}\) is. We actually don’t need it. We can express these ideas in terms of logic and the membership \(\in\) operator. But it is convenient to be able to write things down with custom notation, and as long as we’re careful, this won’t break anything.
We can introduce a notation by introducing an axiom that defines its meaning. For example, for subset, we can introduce the axiom
\[ \forall x (\forall y (x \subseteq y \leftrightarrow \forall z (z \in x \rightarrow z \in y))) \]
This is just defining a new operator in terms of existing operators.
For something like curly brace notation \(\{x,y,z\}\), we can think of it as a variadic operator. Its defining property is
\[ \forall x_1 (\forall x_2 (\ldots \forall x_n (\forall y (y \in \{x_1,x_2,\ldots,x_n\} \leftrightarrow y = x_1 \vee y = x_2 \vee \ldots \vee y = x_n)))) \]
We can’t write a single axiom for this since we want an axiom for each \(n\). So we can add a rule instead like for the axiom schema of specification.
This is an example of constructive notation. Unlike \(\subseteq\), which is just shorthand for a formula, this is shorthand for an object. Another example of constructive notation is binary set intersection \(x \cap y\). We can define it like this:
\[ \forall x (\forall y (\forall z (z \in x \cap y \leftrightarrow z \in x \wedge z \in y))) \]
We have to be careful when we introduce constructive notation like this. Notation should just be shorthand. It shouldn’t allow us to prove anything that we couldn’t prove before. So if you introduce notation that is shorthand for an object with some property, you should first prove that an object with that property exists without using the notation. In math, we call this a conservative extension of our theory. For set intersection, we should prove
\[ \forall x (\forall y (\exists w (\forall z (z \in w \leftrightarrow z \in x \wedge z \in y)))) \]
We can prove this using the axiom schema of specification, as we did above.
Unrestricted set comprehension is an example of a notational definition that is not a conservative extension:
\[ \forall x (x \in \{x : P\} \leftrightarrow P) \]
Where \(x\) occurs free in \(P\). This constructs the set of all objects that satisfy some property \(P\). To see why this is a problem, let’s construct the set
\[ \{x : \neg (x \in x)\} \]
This is the set of all sets that do not contain themselves, whose existence leads to Russel’s paradox. We took great care to prevent us from being able to prove the existence of this set because it leads to a contradiction. Without this notation, it would be impossible to prove this set’s existence. But now, it can be proven. This is an example of why we should make sure our notational definitions are conservative extensions.
There is another, more artificial problem with that notation: In the previous operators, in order to compute free variables and substitution, we can just recur on the subformulas. But here, if we tried to substitute something for \(x\) in a comprehension, we wouldn’t want the \(x\) to be replaced. This notation is like a quantification over x. Are we going to have to specify free variables and substitution on every operator we define? Let’s keep things simple for now and restrict ourselves to operators that don’t bind any variables. It is possible to allow for this kind of notation, but it will complicate the implementation.
Before we move on, let’s take a step back and think about what we did to express set theory in our system. We wrote down some axioms and defined some operators. We also added some rules. Some of these axioms were fundamental, like the axiom of empty set. Others were conservative extensions, like the definition of the subset operator \(\subseteq\) and curly braces. This is how we will express theories in our system. A theory is just a collection of axioms, operators, and rules.
3.2 Natural Numbers
We talked about sets, now let’s finally talk about numbers! Specifically, natural numbers. That is, \(0,1,2,3,\ldots\).
We’ll start out by stating that the constant \(zero\) is a natural number. To do that, we’ll introduce a new operator, \(N\). This is a unary operator, which means it has one argument. \(N(x)\) means that \(x\) is a natural number.
Our first axiom states that 0 is a natural number:
\[N(zero)\]
We are defining the number zero to be the free variable \(zero\).
So far so good. We will also introduce a successor operation, \(S\), which is another unary operator. \(S(x)\) is the successor of \(x\). We define the number 1 to be the successor of 0, \(S(zero)\). And we define 2 to be the successor of 1, \(S(S(zero))\). And so on. The successor operation behaves like adding 1.
Our second axiom states that the natural numbers are closed under succession:
\[\forall n (N(n) \rightarrow N(S(n)))\]
If \(n\) is a natural number, then its successor is a natural number.
Our third axiom tells us about equality of natural numbers:
\[\forall n (\forall m (N(n) \wedge N(m) \rightarrow (S(n) = S(m) \rightarrow n = m)))\]
If the successors of two natural numbers are equal, then the two numbers are equal.
We also state that 0 is not the successor of any natural number:
\[\forall n (N(n) \rightarrow \neg (S(n) = zero))\]
Finally, we have the axiom schema of induction:
Let \(P\) be some formula where \(n\) does not occur free.
\[[P[zero \mathbin{/} x] \wedge \forall n (P[n \mathbin{/} x] \rightarrow P[S(n) \mathbin{/} x])] \rightarrow \forall x (N(x) \rightarrow P)\]
If \(P\) is true for 0 and, for all natural numbers \(n\), \(P\) being true for \(n\) implies that \(P\) is true for \(S(n)\), then \(P\) is true for all natural numbers. This is how we prove that things are true for all natural numbers. Here is the rule:
\[ \frac{ \Gamma \vdash P[zero \mathbin{/} x] \qquad \Gamma, N(n), P[n \mathbin{/} x] \vdash P[S(n) \mathbin{/} x] }{ \Gamma \vdash \forall x (N(x) \rightarrow P) } (NatInd) \]
With the restriction that \(n\) does not occur free in \(P\). This form follows from the axiom schema if we apply some rules.
These are the peano axioms for natural numbers. Normally, people also include axioms that assert reflexive, symmetric, and transitive equality, and that natural numbers are closed under equality \(\forall n (\forall a ((N(n) \wedge n = a) \rightarrow N(a)))\), but these all follow from our definition of equality, so we don’t need these axioms.
We can define addition and multiplication by adding two operators \(+,\cdot\) and some axioms:
\[\forall a (N(a) \rightarrow a + 0 = a)\] \[\forall a (\forall b(N(a) \wedge N(b) \rightarrow a + S(b) = S(a + b)))\] \[\forall a (N(a) \rightarrow a \cdot 0 = 0)\] \[\forall a (\forall b(N(a) \wedge N(b) \rightarrow a \cdot S(b) = a + a \cdot b))\]
For some intuition, let’s look at some examples:
\[a + S(S(S(zero)))\] \[S(a + S(S(zero)))\] \[S(S(a + S(zero)))\] \[S(S(S(a + zero)))\] \[S(S(S(a)))\]
\(a + b\) just applies \(S\) to \(a\) \(b\) times. So addition is repeated succession.
\[a \cdot S(S(S(zero)))\] \[a + a \cdot S(S(zero))\] \[a + a + a \cdot S(zero)\] \[a + a + a + a \cdot zero\] \[a + a + a + zero\] \[a + a + a\]
\(a \cdot b\) just adds \(a\) to itself \(b\) times. So multiplication is repeated addition.
If you want some practice, prove that the natural numbers are closed under addition: \(\forall a(N(a) \rightarrow \forall b (N(b) \rightarrow N(a + b)))\). Hint: you only need to do induction on \(b\).
Now we see how we can use our system to talk about pretty much any area of math. In this post, we built a foundation for mathematics where we can make statements and prove them. In the next part, we will implement this system in code so we can have a computer automatically check our proofs.
4 Extras
Here is some extra exploration of the concepts in this post:
4.1 Natural Numbers as Sets
We can actually define natural numbers in terms of sets. And all we need to do is add a few axioms:
\[ zero = \{\} \]
\[ \forall n (S(n) = \bigcup \{n, \{n\}\}) \]
Where \(\bigcup\) denotes the union of a set’s elements, like in the axiom of union. To define this operator, we’ll add the axiom
\[ \forall F (\forall x (x \in \bigcup F \leftrightarrow \exists Y (x \in Y \wedge Y \in F))) \]
This means that
\[ 0 = \{\} \]
\[ 1 = \{\{\}\} = \{0\} \]
\[ 2 = \{\{\}, \{\{\}\}\} = \{0,1\} \]
\[ 3 = \{\{\}, \{\{\}\}, \{\{\}, \{\{\}\}\}\} = \{0,1,2\} \]
Each natural number is the set of all lesser natural numbers.
Rather than \(S\) being a primitive operator, it is now defined in terms of union and our curly brace notation, which is defined in terms of logic and set membership.
We can also define \(N\) in terms of sets:
\[\forall n (N(n) \leftrightarrow n = \{\} \vee \exists m (N(m) \wedge n = S(m)))\]
This definition is equivalent to the original definition under our new axioms. (We need to use induction to prove the equivalence though).
Now that we have notation, we can more easily express the axiom of infinity, one of the axioms we left out of our set theory:
\[ \exists X (\{\} \in X \wedge \forall y (y \in X \rightarrow S(y) \in X)) \]
This axiom guarantees the existence of an infinite set. You might be surprised to see \(S\) show up in there, but now it’s defined in terms of sets, so it is valid to use in set theory. But this looks like it’s saying something about natural numbers. We can also see that zero, which we have now defined as \(\{\}\), shows up too. In fact, this set is a superset of the natural numbers! We can prove that all natural numbers are in this set by induction. To give an idea of the proof, 0 is in the set by the first condition \(\{\} \in X\) and the second condition is basically an inductive case.
We can then prove the existence of the set of natural numbers \(\mathbb{N}\) using specification and the axiom of infinity:
\[ \exists \mathbb{N} (\forall n (n \in \mathbb{N} \leftrightarrow (n \in X \wedge N(n)))) \]
Where \(X\) is the set from the axiom of infinity.
This, combined with the fact that all natural numbers are in \(X\), allows us to add the axiom
\[ \forall n (n \in \mathbb{N} \leftrightarrow N(n)) \]
Another thing we get for free by defining things in terms of set theory is comparison \(<\). Our definition of natural numbers leads to the fact that every natural number is the set of all natural numbers less than itself. This means we can define natural number comparison in terms of set membership:
\[ \forall a (\forall b (N(a) \wedge N(b) \rightarrow (a < b \leftrightarrow a \in b))) \]
Another consequence of using set theory is that we can get rid of some of our axioms because they follow from our definitions and the axioms of set theory. We can get rid of the axioms that state zero is natural, the successor of a natural is a natural, successors being equal implies that the two numbers are equal (we need another set theory axiom for this, the axiom of regularity), and zero is not the successor of any natural. This means we only need the axioms that define \(N,+,\cdot\) and the axiom schema of induction.
Here is the axiom of regularity that I mentioned:
\[ \forall x (\neg (x = \emptyset) \rightarrow \exists y (y \in x \wedge y \cap x = \emptyset)) \]
This means that every nonempty set has an element that doesn’t overlap with itself. This, along with the axiom of pairing, guarantees that no set is an element of itself.
Set theory is pretty cool! In fact, pretty much all modern mathematics can be expressed in terms of set theory. Real numbers can be defined in terms of rationals, which can be defined in terms of integers, which can be defined in terms of naturals, which can be defined in terms of sets. In geometry, shapes and lines can be defined in terms of sets of points in space. Points in space can be defined in terms of ordered pairs of real numbers, which can be defined in terms of sets. This is why people say that logic and set theory are the foundation of mathematics.
You might be wondering why there is no predicate for sets like how we have \(N\) for naturals. In our set theory, we pretty much assume that everything is a set and that sets just contain other sets. These are called pure sets. You can also have a set theory that permits things that aren’t sets to be elements of sets. If you want to learn more, look up urelements and ZermeloFraenkel set theory with atoms. In practice, since pretty much everything can be defined in terms of sets, pure sets are usually all you need!
4.2 Law of Excluded Middle
These rules we wrote are an adaptation of the system LK. Our adaptation has a few problems though. In LK, in the same way that we have a list of assumptions that are conjoined (and) on the left of the turnstile, we have a list of statements we’re trying to prove on the right side that are disjoined (or). In this system, judgements look like
\[ P_1,P_2,\ldots,P_n \vdash Q_1,Q_2,\ldots,Q_m \]
Which means assuming \(P_1,P_2,\ldots,P_n\), at least one of \(Q_1,Q_2,\ldots,Q_m\) is true. Our judgements are special cases where \(m=1\). In LK, there are some things that you can prove that you cannot prove with our rules. For example, the law of excluded middle:
\[\vdash x \vee \neg x\]
In order to be complete with respect to LK, we need to modify some of our rules and add some more structural rules:
\[ \frac{\Gamma, P \vdash Q \vee R}{\Gamma \vdash (P \rightarrow Q) \vee R} ({\rightarrow}R^*) \] \[ \frac{\Gamma, P \vdash Q}{\Gamma \vdash \neg P \vee Q} ({\neg}R^*) \]
This makes sense since \((P \rightarrow Q) \leftrightarrow (\neg P \vee Q)\).
\[ \frac{\Gamma \vdash P[y \mathbin{/} x] \vee Q}{\Gamma \vdash \forall x (P) \vee Q} ({\forall}R^*) \] Where \(y\) does not occur free in \(P\). \[ \frac{\Gamma \vdash P \vee P}{\Gamma \vdash P} (CR) \] \[ \frac{\Gamma \vdash Q \vee P}{\Gamma \vdash P \vee Q} (PR) \]
With these rules, we are now complete with respect to LK. We can keep our old versions of the rules since they follow from the new versions. The new versions have stars next to their names so we can tell the difference. Here is the proof of the law of excluded middle:
\[ \displaystyle \frac{ \displaystyle \frac{ \displaystyle \frac{ \displaystyle \frac{}{ \displaystyle y \vdash y } (I) }{ \displaystyle \vdash \neg y \vee y } ({\neg}R^*) }{ \displaystyle \vdash y \vee \neg y } (PR) }{ \displaystyle \vdash \forall x (x \vee \neg x) } ({\forall}R) \]
Now that we can prove this, we can create the assumption \(P \vee \neg P\) "out of thin air" using \(Cut\). As an example of why this can be useful, if we’re trying to prove something about natural numbers, we can create the assumption that the number is either even or not even (which implies that it’s odd) and then handle both cases. Sometimes being able to do this makes a proof easier.
4.3 All you need is \(\forall\) and \(\rightarrow\)
All you actually need for our language of statements is \(\forall\) and \(\rightarrow\) (and operators like \(\in\) for theories). These equivalences show how our connectives can be defined in terms of these two things:
\[(\neg p) \leftrightarrow (p \rightarrow \forall x (x))\]
Let’s think about why these two formulas are equivalent. When \(\neg P\) is true, \(P \rightarrow \forall x (x)\) is also true. If we think about proving \(\neg P \rightarrow (P \rightarrow \forall x (x))\), we’d end up with assumptions \(\neg P\) and \(P\). We can use this to prove anything, including \(\forall x (x)\). When \(P \rightarrow \forall x (x)\) is true, \(P\) must be false. If \(P\) was true, we’d be able to prove \(\forall x (x)\), which is a contradiction. If we think about proving \((P \rightarrow \forall x (x)) \rightarrow \neg P\), we’d end up with assumptions \(P\) and \(P \rightarrow \forall x (x)\) and having to prove \(\forall x (x)\), which can be done with \(({\rightarrow}L)\).
\[\exists x (P) \leftrightarrow (\neg \forall x (\neg P)) \leftrightarrow ((\forall x (P \rightarrow \forall x (x))) \rightarrow \forall x (x))\]
To understand this, let’s think about \(\neg \exists x (P)\). This is saying that there does not exist an object with property \(P\). This is the same as saying that all objects don’t have property \(P\), which can also be written as \(\forall x (\neg p)\). And the negation of that is equivalent to \(\neg \neg \exists x (P)\), which is just \(\exists x (P)\).
\[(P \vee Q) \leftrightarrow (\neg P \rightarrow Q) \leftrightarrow ((P \rightarrow \forall x (x)) \rightarrow Q)\]
To see why these two are equivalent, let’s think about both directions. If \(P \vee Q\) is true, then either \(P\) is true or \(Q\) is true (or both). If \(P\) is true, then we can prove \(\neg P \rightarrow Q\) by reaching a contradiction. If \(Q\) is true, we don’t even need to use the \(\neg P\) assumption to prove the \(Q\). In the other direction, let’s assume \(\neg P \rightarrow Q\) is true. We know that either \(P \vee \neg P\) is true by the law of excluded middle. If \(P\) is true, then we can prove \(P \vee Q\) with \(P\). If it is false, we can use the implication and the negation to get \(Q\).
\[(P \wedge Q) \leftrightarrow (\neg (\neg P \vee \neg Q)) \leftrightarrow ((P \rightarrow (Q \rightarrow \forall x (x))) \rightarrow \forall x (x))\]
Let’s think about \(\neg (\neg P \vee \neg Q)\). \(\neg P \vee \neg Q\) is true when either \(P\) or \(Q\) is false (or both). It is only false when neither are false. So \(\neg (\neg P \vee \neg Q)\) is true when neither \(P\) nor \(Q\) are false, which means they’re both true \(P \wedge Q\). This is one of DeMorgan’s laws. The other is the dual: \(P \vee Q \leftrightarrow \neg (\neg P \wedge \neg Q)\).
Negation is a little less weird in this system. \(({\neg}L)\) makes more sense when you think about it in terms of \(({\rightarrow}L)\). We prove \(p\) to reach a contradiction and then we get \(\forall x (x)\) as an assumption, which lets us prove anything.
Pretty cool.