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---
language: en
license: mit
---

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## Dataset card for math_stratos_scale_judged_and_annotated_with_difficulty_incorrect_and_hardest_reannnotated_2

This dataset was made with [Curator](https://github.com/bespokelabsai/curator/).

## Dataset details

A sample from the dataset:

```python
{
    "problem": "Let \\( F(x) \\) be the antiderivative of the function \\( f(x) \\) on the interval \\([a, b]\\), and let \\( \\varphi(y) \\) be a differentiable function on the interval \\([p, q]\\). Suppose \\( a \\leq \\varphi(y) \\leq b \\) for all \\( y \\) in the interval \\([p, q]\\), and for any point \\( y_0 \\) in the interval \\([p, q]\\), there exists a neighborhood \\( U(y_0) \\) such that if \\( y \\in U(y_0) \\) and \\( y \\neq y_0 \\), then \\( \\varphi(y) \\neq \\varphi(y_0) \\). Prove that \\( F(\\varphi(y)) \\) is the antiderivative of the function \\( f(\\varphi(y)) \\varphi'(y) \\).",
    "deepseek_reasoning": "\nOkay, so I need to prove that \\( F(\\varphi(y)) \\) is the antiderivative of \\( f(\\varphi(y)) \\varphi'(y) \\). Let's start by recalling what an antiderivative is. An antiderivative of a function \\( g(y) \\) is a function \\( G(y) \\) such that \\( G'(y) = g(y) \\). So here, I need to show that if I take the derivative of \\( F(\\varphi(y)) \\) with respect to \\( y \\), I get \\( f(\\varphi(y)) \\varphi'(y) \\).\n\nWait, but that sounds like the chain rule. The chain rule states that if you have a composite function \\( F(\\varphi(y)) \\), then its derivative is \\( F'(\\varphi(y)) \\cdot \\varphi'(y) \\). But \\( F \\) is an antiderivative of \\( f \\), so \\( F'(x) = f(x) \\). Therefore, substituting \\( x = \\varphi(y) \\), we have \\( F'(\\varphi(y)) = f(\\varphi(y)) \\). Multiplying by \\( \\varphi'(y) \\) gives the derivative \\( f(\\varphi(y)) \\cdot \\varphi'(y) \\). \n\nTherefore, by the chain rule, the derivative of \\( F(\\varphi(y)) \\) is indeed \\( f(\\varphi(y)) \\varphi'(y) \\), which would mean that \\( F(\\varphi(y)) \\) is an antiderivative of that product. \n\nBut hold on, the problem mentions some conditions about \\( \\varphi(y) \\). It says \\( \\varphi(y) \\) is differentiable on \\([p, q]\\), maps into \\([a, b]\\), and for any point \\( y_0 \\) in \\([p, q]\\), there's a neighborhood around \\( y_0 \\) where \\( \\varphi(y) \\neq \\varphi(y_0) \\) for \\( y \\neq y_0 \\). \n\nI need to make sure these conditions are necessary for the proof. Let me think. The chain rule requires that both the outer function \\( F \\) and the inner function \\( \\varphi(y) \\) are differentiable. Since \\( F \\) is an antiderivative of \\( f \\), it's differentiable (because by definition an antiderivative is differentiable, and its derivative is the original function). Also, \\( \\varphi(y) \\) is given as differentiable on \\([p, q]\\). So the differentiability condition is satisfied. \n\nBut what's the deal with that condition about \\( \\varphi(y) \\neq \\varphi(y_0) \\) in some neighborhood around each \\( y_0 \\)? That seems to ensure that \\( \\varphi \\) is not constant in any neighborhood around a point. If \\( \\varphi \\) were constant in some neighborhood, then its derivative would be zero there, but perhaps the condition allows us to avoid cases where \\( \\varphi'(y) \\) is zero? Wait, but even if \\( \\varphi \\) is constant in some interval, the derivative \\( \\varphi'(y) \\) would be zero there, but \\( F(\\varphi(y)) \\) would also be constant, so its derivative is zero, which matches \\( f(\\varphi(y)) \\varphi'(y) \\), since \\( \\varphi'(y) \\) is zero. \n\nHmm, maybe the condition is there to prevent issues when applying the chain rule in situations where the inner function is not locally injective? Wait, the chain rule doesn't require the function to be injective, as far as I know. Maybe if the inner function has a derivative at a point, even if it's oscillating or something, the chain rule still holds. Let me check.\n\nWait, actually, the chain rule requires that the inner function is differentiable at \\( y_0 \\) and the outer function is differentiable at the inner function's value at \\( y_0 \\). As long as those are satisfied, the chain rule holds. Here, \\( F \\) is differentiable everywhere on \\([a, b]\\), and \\( \\varphi \\) is differentiable on \\([p, q]\\), so the chain rule should hold regardless of the other condition. So perhaps that extra condition is there to ensure that when we talk about \\( F(\\varphi(y)) \\) being an antiderivative, maybe there's some local behavior required.\n\nWait, maybe the problem is assuming potentially that \\( \\varphi(y) \\) might have intervals where it's constant, but given the condition that in any neighborhood around \\( y_0 \\), \\( \\varphi(y) \\neq \\varphi(y_0) \\) except at \\( y_0 \\), it implies that \\( \\varphi \\) is locally injective around each point. That is, \\( \\varphi \\) doesn't stay constant in any neighborhood. Therefore, maybe if \\( \\varphi \\) was constant on some interval, while \\( F(\\varphi(y)) \\) would still be differentiable with derivative zero (since it's constant), the function \\( f(\\varphi(y)) \\varphi'(y) \\) would also be zero there. So perhaps even if \\( \\varphi \\) is constant on some interval, the derivative would agree. So why the condition?\n\nAlternatively, maybe the problem is avoiding the case where \\( \\varphi \\) is constant in a neighborhood, but even so, the chain rule still applies. Maybe that condition is just added to make sure that \\( \\varphi \\) is not a constant function? Or to ensure some form of non-degeneracy?\n\nAlternatively, is the problem perhaps dealing with the inverse function theorem? But we're not inverting here. Maybe it's just that in the hypothesis, if \\( \\varphi(y) \\) is such that around every point \\( y_0 \\), \\( \\varphi(y) \\) is not equal to \\( \\varphi(y_0) \\) when \\( y \\neq y_0 \\), then this ensures that \\( \\varphi'(y) \\) is not locally zero everywhere, so maybe it's a way to ensure that \\( \\varphi \\) is not constant? Wait, no. If \\( \\varphi \\) is not constant, there could still be points where \\( \\varphi'(y) = 0 \\).\n\nHmm, perhaps even this condition is more restrictive. It says for each \\( y_0 \\) in \\([p, q]\\) there is a neighborhood \\( U(y_0) \\) where \\( y \\neq y_0 \\) implies \\( \\varphi(y) \\neq \\varphi(y_0) \\). This actually says that \\( \\varphi \\) is locally injective around each point. Because in some neighborhood, it doesn't take the same value as \\( \\varphi(y_0) \\) except at \\( y_0 \\). So, \\( \\varphi \\) is injective on each such neighborhood.\n\nBut the problem is about proving \\( F(\\varphi(y)) \\) is the antiderivative of \\( f(\\varphi(y))\\varphi'(y) \\). So, unless there's some case where even if the chain rule holds, \\( F(\\varphi(y)) \\) might not be an antiderivative? Wait, maybe in the case where \\( \\varphi \\) is not injective, integrating \\( f(\\varphi(y))\\varphi'(y) \\) might require considering overlapping intervals or something else? But no, from calculus, we know that the integral of \\( f(\\varphi(y))\\varphi'(y) dy \\) is \\( F(\\varphi(y)) + C \\), regardless of the properties of \\( \\varphi \\), as long as \\( \\varphi \\) is differentiable and \\( f \\) is continuous. Wait, but in the problem statement, \\( F \\) is just an antiderivative of \\( f \\), which requires that \\( f \\) is Riemann integrable? Wait, or perhaps \\( f \\) is continuous? If \\( F \\) is the antiderivative of \\( f \\), then \\( F' = f \\), so \\( f \\) must be continuous if \\( F \\) is continuously differentiable? Wait, not necessarily; the derivative of a differentiable function isn't necessarily continuous. So, \\( F \\) is differentiable, and \\( f \\) is its derivative, which may not be continuous.\n\nBut for the chain rule to hold, we need the outer function (here \\( F \\)) to be differentiable at the inner function's (here \\( \\varphi(y) \\)) point. Since \\( F \\) is differentiable everywhere on \\([a,b]\\) because it's the antiderivative of \\( f \\) on \\([a, b]\\), and \\( \\varphi(y) \\) is in \\([a, b]\\) for all \\( y \\in [p, q]\\), then \\( F \\) is differentiable at all points \\( \\varphi(y) \\), so the chain rule applies.\n\nBut the problem's given conditions might be about \\( \\varphi \\) not being constant in any neighborhood. Maybe that's just to ensure that the derivative \\( \\varphi'(y) \\) isn't zero everywhere in a neighborhood. Wait, but in a neighborhood where \\( \\varphi \\) is constant, the derivative would be zero, but then the composition \\( F(\\varphi(y)) \\) is also constant, so its derivative is zero, and \\( f(\\varphi(y))\\varphi'(y) \\) is also zero, since \\( \\varphi'(y) \\) is zero. So even in that case, the equality holds. So the chain rule holds regardless.\n\nTherefore, why is the condition given that in any neighborhood of \\( y_0 \\), \\( \\varphi(y) \\neq \\varphi(y_0) \\) for \\( y \\neq y_0 \\)? \n\nPerhaps to make sure that \\( \\varphi \\) is a function where every point is either a point where the function is strictly increasing or decreasing nearby, so that \\( \\varphi \\) doesn't have any intervals where it's flat? But I still don't see how that affects the derivative. Let me check an example.\n\nSuppose \\( \\varphi(y) \\) is a constant function. Then \\( \\varphi'(y) = 0 \\), \\( F(\\varphi(y)) \\) is a constant, so derivative is 0, and \\( f(\\varphi(y))\\varphi'(y) \\) is also 0. So that works. Now suppose \\( \\varphi(y) \\) is not injective but has some points where \\( \\varphi'(y) = 0 \\). For example, \\( \\varphi(y) = y^2 \\) on some interval including 0. At y=0, the derivative is 0, and around 0, the function isn't injective (since it's decreasing on the left and increasing on the right). The problem's condition that in any neighborhood of y0, \u03c6(y) \u2260 \u03c6(y0) for y \u2260 y0. But if \u03c6(y) = y^2 and y0=0, in any neighborhood of 0, there are points y\u22600 where \u03c6(y)=\u03c6(0)=0 only at y=0, which isn't the case. Wait, \u03c6(y)= y^2, then \u03c6(y)=0 only at y=0. So in this case, even though \u03c6(y) is not injective around 0, but for any neighborhood U(0), if y\u22600 is in U(0), \u03c6(y) = y2 \u22600. So that satisfies the condition. Hmm, so this \u03c6(y) would satisfy the given conditions.\n\nBut if \u03c6(y) is, say, a constant function on some interval, then for any y0 in that interval, in the neighborhood around y0, \u03c6(y) is equal to \u03c6(y0) for all y in that interval. Therefore, that does not satisfy the given condition (since in any neighborhood of y0, there's y \u2260 y0 with \u03c6(y) = \u03c6(y0)). Therefore, such a \u03c6(y) would not satisfy the condition. So the condition essentially rules out \u03c6(y) being constant on any interval. That is, \u03c6(y) can't be locally constant anywhere. So \u03c6(y) has to be such that around every point y0, \u03c6(y) is not constant.\n\nTherefore, if \u03c6(y) is such that around every y0, \u03c6(y) is different from \u03c6(y0) when y \u2260 y0, then \u03c6 is \"locally non-constant\" everywhere. That might prevent \u03c6 from having intervals where it's flat, thus ensuring that the derivative \u03c6\u2019 isn't zero over an interval. But even then, even a function that is not constant can have isolated points where the derivative is zero.\n\nSo, maybe the function \u03c6 is \"non-constant\" in every neighborhood, but that might just mean that \u03c6 is not constant in any interval, but it could still have points where the derivative is zero, just not over an interval. For example, \u03c6(y) = y + sin(1/y) near y=0 (but adjusted to be smooth). Such a function could oscillate but not be constant in any neighborhood of 0. However, the problem's condition would require that for each y0, there exists a neighborhood where \u03c6(y) \u2260 \u03c6(y0) for y \u2260 y0. But such a function as \u03c6(y) = y + sin(1/y) isn't injective in any neighborhood around 0. Therefore, the condition given is stronger: in a sense, \u03c6 is injective on some neighborhood around each point. Because in the given condition, for each y0, there exists a neighborhood where \u03c6(y) is injective (since \u03c6(y) \u2260 \u03c6(y0) for y \u2260 y0). Wait, actually, that's not quite injective. It only says that \u03c6(y0) is not taken by any other y in the neighborhood, but it does not say that \u03c6(y1) \u2260 \u03c6(y2) for y1 \u2260 y2 in that neighborhood. So it's a weaker condition than local injectivity. It's only that \u03c6(y) doesn't attain the value \u03c6(y0) again in that neighborhood.\n\nOkay, so in other words, around every y0, \u03c6(y) is \"isolated\" in the sense that \u03c6(y0) is not attained by any other point in that neighborhood. That does not prevent \u03c6 from taking other values multiple times. Wait, actually, a function with this property is called \"locally injective\" only at points where the function is injective in a neighborhood. But the problem states a weaker condition. For every y0, there exists a neighborhood U(y0) such that for all y in U(y0), y \u2260 y0 \u21d2 \u03c6(y) \u2260 \u03c6(y0). So this is different from local injectivity. For example, consider \u03c6(y) such that \u03c6(y) = y for y < 0 and \u03c6(y) = -y for y \u2265 0. Then at y0=0, take any neighborhood around 0. For y \u2260 0, \u03c6(y) = -y if y > 0, so \u03c6(y) in (0, \u03b5) is (-\u03b5, 0), and \u03c6(y) in (-\u03b5, 0) is (-\u03b5, 0) as well. Therefore, \u03c6(y)=0 only at y=0. So in any neighborhood around 0, \u03c6(y) does not repeat \u03c6(0)=0. Similarly, for other points y0\u22600, say y0 > 0, in a small neighborhood around y0, \u03c6(y) = -y, which is injective there. So, in this case, \u03c6 satisfies the given condition but is not injective globally. \n\nBut, on the other hand, \u03c6 could be a function that \"folds\" points infinitely often but in such a way that each value is taken only once in each neighborhood. Wait, maybe. For example, \u03c6(y) = y sin(1/y) near y=0 (but maybe adjusted to avoid oscillation). Wait, no, if \u03c6(y) = y sin(1/y) for y\u22600 and \u03c6(0)=0. Then, near y=0, for any neighborhood around 0, there are y\u22600 such that \u03c6(y)=0, which is the same as \u03c6(0). So this function wouldn't satisfy the condition.\n\nBut if the function is like \u03c6(y) = y for y \u22600 and \u03c6(0)=0. Then, trivially, it still satisfies the condition. Hmm.\n\nAlternatively, perhaps the role of that condition is to make sure that when we take the derivative of F(\u03c6(y)), the function \u03c6(y) isn't \"stationary\" at \u03c6(y0) for intervals around y0, which could affect the differentiation. Wait, but differentiation is a local property at the point y0. The chain rule doesn't care about the behavior of \u03c6 away from y0 except that \u03c6 is differentiable at y0.\n\nBut perhaps in some generalized sense, if the function \u03c6(y) were to be constant in a neighborhood, then F(\u03c6(y)) would be constant there, and the derivative would be zero, which is equal to f(\u03c6(y)) \u03c6\u2019(y). So even then, everything works. So why is the condition imposed?\n\nAlternatively, maybe the problem is not in the differentiation step but in the initial setup where F is an antiderivative. If \u03c6(y) were not satisfying that condition, could there be an issue with composing F and \u03c6?\n\nWait, perhaps if \u03c6(y) has a local maximum or minimum at some point y0, then in a neighborhood of y0, \u03c6(y) could be equal to \u03c6(y0) on one side. For example, take \u03c6(y) = (y - y0)^2, then in any neighborhood around y0, \u03c6(y) = (y - y0)^2, so \u03c6(y) attains the value \u03c6(y1) for y1 \u2260 y0. Wait, but in this case, \u03c6(y1) is different from \u03c6(y0) when y1 \u2260 y0. Wait, if y0 is part of an extremum, say, \u03c6(y) = (y - y0)^2 + c, then in a neighborhood around y0, \u03c6(y) will attain values greater or equal to c. But if y0 is a minimum, \u03c6(y0)=c, then \u03c6(y) > c for y \u2260 y0 in the neighborhood. So in that case, the condition is satisfied because \u03c6(y) \u2260 \u03c6(y0) for y\u2260y0.\n\nBut in the case of \u03c6(y) being a constant function in a neighborhood around y0, then the condition would not hold. Therefore, this condition is necessary to exclude functions that are constant in neighborhoods. But again, in such a case, as mentioned before, the derivative would still hold. So maybe it's a redundant condition? Or maybe it's included to handle some other aspect?\n\nAlternatively, is the problem possibly related to the inverse function theorem? For example, if \u03c6 is injective in a neighborhood of y0, then we might use differentiability of the inverse or something, but the chain rule doesn't need that.\n\nAlternatively, maybe in higher dimensions or under different conditions, having such a property is crucial, but in single-variable calculus, perhaps not. However, maybe there's an issue when integrating over intervals where \u03c6 might backtrack? But no, because in integrating f(\u03c6(y)) \u03c6\u2019(y) dy over [p, q], even if \u03c6 is not injective, the integral would still evaluate correctly via substitution. As in, the integral would account for multiple passes over the same region in x. Wait, but let me think.\n\nHowever, in this problem, we are just talking about antiderivatives, not definite integrals. So, to define an antiderivative, the chain rule composition should hold everywhere, regardless of the path. Therefore, as long as F is an antiderivative of f, and \u03c6 is differentiable, then the derivative of F(\u03c6(y)) is f(\u03c6(y)) \u03c6\u2019(y) by the chain rule. Therefore, regardless of the other conditions, the statement seems to hold. \n\nBut perhaps the problem adds those conditions to prevent functions \u03c6 where \u03c6'(y) is zero even in neighborhoods but f(\u03c6(y)) could be non-zero. Wait, no. If \u03c6(y) is constant on a neighborhood, then \u03c6\u2019(y) is zero there, and so f(\u03c6(y)) \u03c6\u2019(y) is zero, matching the derivative of F(\u03c6(y)) which is constant there.\n\nThus, perhaps the given conditions are unnecessary for the conclusion. That is, even without the condition about \u03c6(y) \u2260 \u03c6(y0) in a neighborhood, the statement would still hold. However, maybe the problem wants to exclude functions \u03c6 that have intervals where they are constant, thereby possibly leading to F(\u03c6(y)) having intervals of constancy. But again, even with that, the chain rule would hold.\n\nTherefore, why is this condition included in the problem statement? Maybe the problem is set up in such a way where the non-repeating of \u03c6(y0) is required for some other reason. Alternatively, perhaps there's a different approach to the proof that uses this condition, such as a substitution or directly working with the definition of the derivative. Let me consider proving it from the definition of the derivative.\n\nSuppose we want to compute the derivative of F(\u03c6(y)) at a point y0. The derivative is the limit as h approaches 0 of [F(\u03c6(y0 + h)) - F(\u03c6(y0))]/h. If \u03c6(y0 + h) \u2260 \u03c6(y0) for h \u22600 small enough, then [F(\u03c6(y0 + h)) - F(\u03c6(y0))]/h = [F(\u03c6(y0 + h)) - F(\u03c6(y0))]/[\u03c6(y0 + h) - \u03c6(y0)] * [\u03c6(y0 + h) - \u03c6(y0)]/h. Then, as h approaches 0, the first factor [F(\u03c6(y0 + h)) - F(\u03c6(y0))]/[\u03c6(y0 + h) - \u03c6(y0)] approaches F\u2019(\u03c6(y0)) = f(\u03c6(y0)), because \u03c6(y0 + h) - \u03c6(y0) approaches 0 (since \u03c6 is continuous, differentiable implies continuous). And the second factor [\u03c6(y0 + h) - \u03c6(y0)]/h approaches \u03c6\u2019(y0). Thus, the product is f(\u03c6(y0)) \u03c6\u2019(y0).\n\nHowever, this splitting into two fractions is only valid if \u03c6(y0 + h) \u2260 \u03c6(y0) for sufficiently small h \u22600. This is exactly where the given condition comes into play. The condition states that in some neighborhood around y0, \u03c6(y) \u2260 \u03c6(y0) if y \u2260 y0. Therefore, for such \u03c6(y), the difference quotient [F(\u03c6(y)) - F(\u03c6(y0))]/[\u03c6(y) - \u03c6(y0)] is defined for y \u2260 y0 in a neighborhood around y0, so the limit as y approaches y0 can be taken as h\u21920. Therefore, the problem's condition ensures that in the definition of the derivative, we don't have to worry about division by zero in the intermediate step. \n\nAh! Therefore, the condition that \u03c6(y) \u2260 \u03c6(y0) in a neighborhood around y0 (except at y0) guarantees that for y near y0 but not equal to y0, \u03c6(y) \u2260 \u03c6(y0), so that we can use the mean value theorem or express the derivative as the product of the derivatives. So if \u03c6(y) - \u03c6(y0) is non-zero for y \u2260 y0 near y0, then we can write the difference quotient as a product, as above. However, if there are points y arbitrarily close to y0 where \u03c6(y) = \u03c6(y0), then we can't split the difference quotient in this way, and the usual chain rule proof via limits would not directly apply. \n\nTherefore, the problem's condition is essential for allowing the step where we divide by \u03c6(y) - \u03c6(y0) and then multiply by \u03c6(y) - \u03c6(y0)/h. Without that condition, there could be points where \u03c6(y) = \u03c6(y0) for y arbitrarily close to y0, which would make that intermediate step undefined. So to handle those cases, we need to ensure that either \u03c6(y) - \u03c6(y0) \u2260 0 for y \u2260 y0 near y0 or handle the case when it does. However, even if \u03c6(y) - \u03c6(y0) =0 for some y near y0, the derivative could still be calculated as f(\u03c6(y0))\u03c6\u2019(y0). Let me think.\n\nFor example, suppose \u03c6(y) is a function that is constant at \u03c6(y0) for infinitely many points approaching y0. Then, even though \u03c6 is differentiable at y0, maybe we have sequences h_n \u21920 such that \u03c6(y0 + h_n) = \u03c6(y0). For those h_n, the difference quotient [F(\u03c6(y0 + h_n)) - F(\u03c6(y0))]/h_n is 0. Meanwhile, the other part of the derivative, if we consider points where \u03c6(y0 + h_n) \u2260 \u03c6(y0), then the quotient would approach f(\u03c6(y0))\u03c6\u2019(y0). However, if between those sequences, the quotient is both 0 and approaching f\u03c6\u2019 , to have the limit exist, \u03c6\u2019(y0) must be zero. Therefore, in this case, the derivative of F(\u03c6(y)) at y0 would still be f(\u03c6(y0))\u03c6\u2019(y0), even if \u03c6\u2019(y0)=0 and parts of the difference quotient are zero and others approach zero.\n\nBut then, even if \u03c6(y) = \u03c6(y0) for some y near y0, as long as \u03c6 is differentiable at y0, the chain rule still holds. So actually, the chain rule in single-variable calculus does not require the function \u03c6 to be locally injective around y0. That's my confusion.\n\nWait, the standard chain rule proof goes like this. Suppose \u03c6 is differentiable at y0, and F is differentiable at \u03c6(y0). Let\u2019s write the difference quotient:\n\n[F(\u03c6(y0 + h)) - F(\u03c6(y0))]/h = [F(\u03c6(y0 + h)) - F(\u03c6(y0))]/[\u03c6(y0 + h) - \u03c6(y0)] * [\u03c6(y0 + h) - \u03c6(y0)]/h\n\nNow, even if \u03c6(y0 + h) - \u03c6(y0) is zero for some h, the first factor [F(\u03c6(y0 + h)) - F(\u03c6(y0))]/[\u03c6(y0 + h) - \u03c6(y0)] could be undefined. However, if \u03c6\u2019(y0) \u2260 0, then for sufficiently small h, \u03c6(y0 + h) - \u03c6(y0) \u2248 \u03c6\u2019(y0) h, so if \u03c6\u2019(y0) \u22600, then \u03c6(y0 + h) \u2260 \u03c6(y0) for small h\u22600, hence the first factor is defined. In that case, the chain rule holds as the limit becomes the product. If \u03c6\u2019(y0) = 0, then  h \u03c6\u2019(y0) =0, so even if there are some h for which \u03c6(y0 + h) = \u03c6(y0), the first factor might not be defined, but since \u03c6\u2019(y0)=0, in the limit the derivative is F\u2019(\u03c6(y0)) * 0 = 0. So the difference quotient is either approaching F\u2019(\u03c6(y0)) * \u03c6\u2019(y0) in the normal case or in the case where we have points where \u03c6(y0 + h) = \u03c6(y0), then the difference quotient is zero for those h, but since in the limit h approaches zero, we have to argue that all approaches give the same limit. But \u03c6\u2019(y0) is zero, so even if there are sequences where the quotient is zero, and others where it is F\u2019(\u03c6(y0)) * \u03c6\u2019(y0), which is zero. So even if F is differentiable, and \u03c6 is differentiable, even with points where \u03c6(y0 + h) = \u03c6(y0), the chain rule holds. Therefore, the chain rule in standard calculus does not require the injectivity condition.\n\nThus, coming back to the original problem, the condition they added might be superfluous? But since the problem includes that condition, maybe the original intent is to use the condition to split the difference quotient as per the earlier method. However, given the standard chain rule does not require this condition, but on the other hand, maybe in some contexts where the derivative is defined differently or if considering more pathological functions, such a condition is needed. \n\nGiven that the problem statement includes the condition, even if in standard calculus it might not be necessary, we should probably use it in the proof. So here's how.\n\nWe can start by taking an arbitrary y0 in [p, q]. By the given condition, there exists a neighborhood U(y0) where \u03c6(y) \u2260 \u03c6(y0) whenever y \u2260 y0. Therefore, in this neighborhood, for y \u2260 y0, \u03c6(y) - \u03c6(y0) \u2260 0. Therefore, the difference quotient [F(\u03c6(y)) - F(\u03c6(y0))]/[\u03c6(y) - \u03c6(y0)] is defined. \n\nSo, the derivative of F(\u03c6(y)) at y0 is the limit as y approaches y0 of [F(\u03c6(y)) - F(\u03c6(y0))]/[y - y0] = [F(\u03c6(y)) - F(\u03c6(y0))]/[\u03c6(y) - \u03c6(y0)] * [\u03c6(y) - \u03c6(y0)]/[y - y0] = F\u2019(\u03c6(y1)) * \u03c6\u2019(y0), where y1 is some point between y and y0 (by the Mean Value Theorem applied to F), but since F is differentiable everywhere, the first term approaches F\u2019(\u03c6(y0)) = f(\u03c6(y0)) as y approaches y0. Meanwhile, the second term is the difference quotient for \u03c6(y), which approaches \u03c6\u2019(y0). Therefore, by limit laws, the product of the limits equals the limit of the product, which gives F\u2019(\u03c6(y0)) * \u03c6\u2019(y0) = f(\u03c6(y0)) \u03c6\u2019(y0). \n\nBut the application of the Mean Value Theorem here would require \u03c6 to be continuous and F to be differentiable. However, F is an antiderivative, so differentiable, \u03c6 is differentiable hence continuous. Therefore, perhaps in this argument, since in the neighborhood around y0, \u03c6(y) \u2260 \u03c6(y0) for y\u2260y0, so non-zero denominator. But with standard calculus even without this, the chain rule proof holds as the entire expression is F\u2019(\u03c6(y0))\u03c6\u2019(y0).\n\nWait, but in some generalizations of derivatives, the condition is necessary. But in standard calculus, differentiability is enough. \n\nTherefore, the answer is correct via the chain rule regardless. But the problem gives this condition, so must use it. Maybe they have a different method in mind? Perhaps through substitution.\n\nAnyway, the key is that if F is the antiderivative of f, then d/dy F(\u03c6(y)) = F\u2019(\u03c6(y)) \u03c6\u2019(y) = f(\u03c6(y)) \u03c6\u2019(y). Therefore, F(\u03c6(y)) is the antiderivative of f(\u03c6(y)) \u03c6\u2019(y). The conditions on \u03c6(y) (the local injectivity around each point) ensure that we can apply the chain rule in terms of the difference quotient, because \u03c6(y) - \u03c6(y0) \u22600 in a neighborhood of y0 (except at y0), allowing division by \u03c6(y) - \u03c6(y0). \n\nSo the conclusion is correct, and the problem conditions are to ensure that in a neighborhood of every point y0, \u03c6(y) \u2260 \u03c6(y0) so that when computing the derivative, we don't have division by zero in the intermediate steps. This makes the standard chain rule application valid.\n\nThus, the proof proceeds by using the chain rule, leveraging the given condition to ensure the difference quotient can be suitably split into the product, allowing the limit to be evaluated as the product of the individual derivatives.\n\n**Final Answer**\n\\boxed{F(\\varphi(y))} \\text{ is the antiderivative of } \\boxed{f(\\varphi(y)) \\varphi'(y)}.\n",
    "deepseek_solution": "To prove that \\( F(\\varphi(y)) \\) is the antiderivative of \\( f(\\varphi(y)) \\varphi'(y) \\), we start by noting that \\( F(x) \\) is an antiderivative of \\( f(x) \\), meaning \\( F'(x) = f(x) \\). We need to show that the derivative of \\( F(\\varphi(y)) \\) with respect to \\( y \\) is \\( f(\\varphi(y)) \\varphi'(y) \\).\n\nUsing the chain rule, the derivative of \\( F(\\varphi(y)) \\) with respect to \\( y \\) is:\n\\[\n\\frac{d}{dy} F(\\varphi(y)) = F'(\\varphi(y)) \\cdot \\varphi'(y)\n\\]\nSince \\( F'(\\varphi(y)) = f(\\varphi(y)) \\), this simplifies to:\n\\[\n\\frac{d}{dy} F(\\varphi(y)) = f(\\varphi(y)) \\cdot \\varphi'(y)\n\\]\n\nThe given conditions on \\( \\varphi(y) \\) ensure that for any point \\( y_0 \\in [p, q] \\), there exists a neighborhood \\( U(y_0) \\) such that \\( \\varphi(y) \\neq \\varphi(y_0) \\) for \\( y \\neq y_0 \\). This condition allows us to use the chain rule without division by zero in the difference quotient. Specifically, the difference quotient:\n\\[\n\\frac{F(\\varphi(y)) - F(\\varphi(y_0))}{y - y_0}\n\\]\ncan be rewritten as:\n\\[\n\\frac{F(\\varphi(y)) - F(\\varphi(y_0))}{\\varphi(y) - \\varphi(y_0)} \\cdot \\frac{\\varphi(y) - \\varphi(y_0)}{y - y_0}\n\\]\nfor \\( y \\neq y_0 \\). As \\( y \\to y_0 \\), the first factor approaches \\( F'(\\varphi(y_0)) = f(\\varphi(y_0)) \\) and the second factor approaches \\( \\varphi'(y_0) \\).\n\nThus, the derivative of \\( F(\\varphi(y)) \\) with respect to \\( y \\) is indeed \\( f(\\varphi(y)) \\varphi'(y) \\), proving that \\( F(\\varphi(y)) \\) is the antiderivative of \\( f(\\varphi(y)) \\varphi'(y) \\).\n\n\\[\n\\boxed{F(\\varphi(y))} \\text{ is the antiderivative of } \\boxed{f(\\varphi(y)) \\varphi'(y)}.\n\\]"
}
```

## Loading the dataset

You can load this dataset using the following code:

```python
from datasets import load_dataset

dataset = load_dataset("mlfoundations-dev/math_stratos_scale_judged_and_annotated_with_difficulty_incorrect_and_hardest_reannnotated_2", split="default")
```

