 Let's try to develop some tools for working with the natural numbers. A useful property is cancellation, and we use that every time we do something like the following. If x plus a is equal to y plus a, then x is equal to y. We'll prove this. So what is that? Well, we want to prove that for all x, y, and a natural numbers, we'd like it if x plus a equals y plus a if and only if x is equal to y. Since this is an if and only if statement, this is actually a biconditional, so we need to prove two conditionals, and these are if x plus a equals y plus a, then x equals y, and also if x equals y, then x plus a is equal to y plus a. So let's try to prove our first conditional. So remember, we can always assume the antecedent of a conditional, so suppose x plus a is equal to y plus a, and we want to conclude that x is equal to y, so we'll write it down here and leave some space in between the two to write our proof. And that's kind of a big gap, so let's see if we can prove something maybe a little bit easier. Rather than trying to prove that this is true for any a, it appears we want to prove that if x plus 0 is equal to y plus 0, then x equals y, and if x plus 1 equals y plus 1, then x equals y, and if x plus 2 equals y plus 2, then x equals y, and so on. And this is the type of thing that we would ordinarily prove by induction. So remember a proof by induction, we have to first prove the base step that our statement is true for 0. So we want to prove that if x plus 0 is equal to y plus 0, then x is equal to y. So remember we can always assume the antecedent of a conditional, so we have x plus 0 is equal to y plus 0, and remember our definition of addition tells us what n plus 0 is, and so we know that x is equal to y, and that proves our base step. Now our induction step is a little hard to state, but it's important that we get it right because it'll help structure our proof. So remember we want to prove that if x plus k equals y plus k, then x equals y. And notice that we're actually trying to prove a conditional. Remember the induction step is if our statement is true for k, then it will be true for the next one. So our induction step will be if, if x plus k equals y plus k, then x equals y, then if x plus k star equals y plus k star, then x equals y. Remember the induction step is always if our statement is true for something, then it's true for the next thing. So again remember we can always assume the antecedent of the conditional we're trying to prove, so we can assume that if x plus k equals y plus k, then x equals y, and since we're trying to prove a conditional, we can always assume the antecedent of a conditional. So let x plus k star equals y plus k star. And let's keep our destination in mind. We'd like to conclude that x is equal to y. So let's try to complete our bridge. Since we're dealing with addition, and definitions are the whole of mathematics, all else's commentary, our definition of addition, tells us that this will be the successors of x plus k and y plus k. Now, what do we know about the successors? Those were introduced with the axioms themselves. Strictly speaking, the axioms are not definitions, but for all intents and purposes we can treat them as if they were definitions. And our axiom tells us that if two successors are equal, then the things that they're successors of must also be equal. So we know that x plus k must be y plus k. But we assume that if x plus k equals y plus k, then x is equal to y. And that completes our bridge. And that completes our induction step. If our statement is true for k, then our statement is also true for the successor of k. And since we've proved the base step, we know that for all natural numbers, if x plus k equals y plus k, then x is equal to y. Let's try to go the other way using an induction proof. So if x equals y, definitions are the whole of mathematics, all else is commentary, n is the same as n plus 0. So x is the same as x plus 0, and y is the same as y plus 0. And that gives us our base step. And again, our induction step is a little complicated to state, but we can take it one step at a time since we want to prove that if x equals y, then x plus k equals y plus k. Our induction step will be proving that if this is true for something k, then it's true for the next thing, k star. So again, we can always assume the antecedent of a conditional. So suppose that if x equals y, then x plus k equals y plus k. Again, we can always assume the antecedent of a conditional, x equals y. So the conditional we began with tells us that x plus k is equal to y plus k. Let's go ahead and put our destination down. What we'd like to conclude is that x plus k star is equal to y plus k star. Definitions are the whole of mathematics, all else is commentary. x plus k star is the same as the successor of x plus k. And since this is a definition, that allows us to go back one step and a final link will be through the axioms. For any natural number, the successor does exist and it's unique. And so when we go from x plus k to the successor of x plus k, we're allowed to do that by the axioms. And there's our induction step. If our statement is true for k, then it's also true for k star. And that completes the proof of the second part of the biconditional.