In this session we'll come back to what we did in the last one. Namely, proving equational properties of programs using structural induction. We will practice the newly learned techniques in a somewhat larger proof. For a more difficult example let's consider the reverse function. We pick its inefficient definition here because it's more amenable to equational proofs. So as defining process we would have nil reverses nil and x followed by XS reverse is the reversal of the list xs followed by the element X. These two equations are as we know equivalent to the more efficient version of fork left and closer to what we want to prove. So we pick them which of course does not prevent us from using at one time the more efficient definition of reverse. So what we would like to do is prove the proposition that XS reverse of reverse is XS. So to prove this it's an obvious induction on the list excess, the base case is really easy. Nil reverse, reverse is the same thing as nil reverse by the first law of reverse which says that nil dot reverse is nil and then we again invoke the first law To show that, that expression now is the same as nil, which establishes the proposition. Let's look at the induction step so here we would have X followed by XS, and then a double reverse. What can we do with that? Well, we can apply the second clause of reverse, which would mean that this expression here would, can be rewritten to the right-hand side, excess.reverse followed by X. And then we have our reverse on both sides here. What else can we do? Well, That doesn't seem to be anything obvious. So let's turn to the right hand side. The right hand side would read simply X followed by XS. So what can we do that? Well, one thing we could do is apply the induction hypothesis which says that the list xs is the same as the list xs reverse, reverse. So then we are left with x followed by xs reverse, reverse and again, there's not much we can do anymore to this side here. So both sides, unfortunately, have simplified to different expressions. This expression here and I've that expression there. So we still need to show that the two sides are the same and proving it directly by induction doesn't work as we have seen. What we can try instead is, we can generalize this equation. So the idea is that, instead of just saying excess to dot reverse here and here. We replace that with. An arbitrary list ys. So our new lemma that we want to prove is for, that for any list ys, ys followed by x reverse is the same as x followed by ys.reverse. And to prove that equation, we can use a second induction argument, this time on the list bias so let's try that. So, let's look at the base case first. So here, YS equals nil and the equation we want to show is that nil followed by x.reverse is the same as x followed by nil.reverse. That's the instansation of the lemma that we want to show here. So, what could we do with the left hand side here? Well, By the first close of plus, plus, nil is a left unit, so this thing simply simplifies to list of x.reverse. reverse. Then the next step would be to expand what list of X is. So list of X, as we know by its definition, is X followed by nil. In the next step then, we would involve the second clause of reverse to arrive at this expression here. When actually there's one intermediate step that we have to do here. So by the second clause of reverse, what would we get? We would get the list that follows the head element first so there would be nil.reverse Followed by, this suffix and now we can simply find nil.reverse by the first clause of reverse is just nil. Where as list of x expands to x followed by nil. So, that's how we arrive, at this expression here. Now what we can do here is we can again invoke the first clause of plus, plus to say nil is a left unit and we're just left with x followed by nil. There's one more step to do. Again by the first clause of reverse we know that nil.reverse is nil. Or we've just now used this equation backwards from going from nil to nil.reverse And that's the right hand side if you want to show here. So we get equality here and the base case is established. Let's look at the inductive step. What we need to do here is show that y followed by ys, that's our list, followed by x reverse is the same as x followed by then y followed by ys reversed. So let's see how we would go about that let's work on the left hand side here. First thing we can do is we can pull out the y from its binding with the list YS, using the second list of comcat. So we have the Y as a head element here, and then the list YS, followed by list of X. The next thing we can do is we can invoke the law of reverse, which says well reverse of a list that starts with Y is the same as reversal of the rest of the list here. And the y becomes the last element of the new list. The next thing we can do is apply the induction hypothesis because we see here that the expression ys followed by list x reverse, that's the left hand side of the equation with just the list ys and we can assume that, that, that, that equation holds. So we can rewrite it to the right hand side of the equation which will be x followed by yes.reverse. Now we can apply the first clause of plus, plus to pull out the accelerant. And we can apply the second clause of reverse to establish that Y followed by ys reverse, is the same as ys reverse followed by a list of y. And again we have equality here. So the auxiliary equation is established and because the auxiliary equation was the last thing was needed to prove the main proposition, that accessory versus success, we adapt. So the proof methodology you have seen here, worked in essentially three steps We could apply a defining equation, either of reverse or contact and we could apply it in two different ways. So. What you've seen here, for instance, going from here to here that was, we invoked the equation left to right. The, that was the second clause of plus, plus, which says plus, plus on a list, with a head element y is the same thing as, a list that starts with y. So we can pull out the y. And that step is called typically, an unfold step. The other, equation of resending step was what you've seen here in the last step where we have applied an equation backwards. The equation for reverse right. Y followed by ys reverse is ys reverse followed by this stuff. Y. And that is called a forward step. So what's important here is that these equations are real equations in the mathematical sense. They can be applied both way-, both ways. They're commutative. And the last step that we typically do in an, a proof like that is apply the induction hypothesis. So the proof method that we've seen is sometimes called the fold-unfold method. For equational proofs of functional programs. So, we finish the session with an exercise, which is a bit more involved than the previous ones, and, it's open ended, so I won't give you a solution immediately. So what I want you to do is, prove another law that's useful, that relates map, and concatenation. So the law says that, essentially the, a, a, map distributes over concat. For any lists XS, YS and function F, XS followed by YS and then map the function is the same thing as mapping the function over XS, mapping it over YS, and concatenating the results. What you need for the proofs is the classes of plus, plus, that you've seen, as well as two classes for maps that you see here. So, again, they derive directly from the definition of map. First clause says to map a function over the empty list, you would get the empty list. The second clause says that to map a function over a list consisting of x followed by xs. What you get is f applied to the head element x followed by the result of mapping f over the rest of.
