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  "body": "\n## Proof\n\nWe compute the derivative of $\\delta$ from its difference quotients.\n\n1. The set $I - c$ is an interval, since it is the image of the interval $I$ under the shift $t \\mapsto t - c$.\n2. For $t \\in I - c$ we have $t + c \\in I$. So $\\delta(t) = \\gamma(t + c)$ is defined, and $\\delta(t) \\in U$ because $\\gamma$ takes values in $U$.\n3. Fix $t \\in I - c$. For each real $h \\ne 0$ with $t + h \\in I - c$, we have $\\dfrac{\\delta(t + h) - \\delta(t)}{h} = \\dfrac{\\gamma(t + c + h) - \\gamma(t + c)}{h}$.\n4. As $h \\to 0$, the right side tends to $\\gamma'(t + c)$, because $\\gamma$ is differentiable at $t + c \\in I$. Here the admissible $h$ are the same on both sides, so one-sided derivatives at endpoints also match.\n5. So $\\delta$ is differentiable at $t$ and $\\delta'(t) = \\gamma'(t + c)$.\n6. Since $\\gamma$ is a [[solution|04SW7WECS8]] of the [[autonomous system|07BNRFQ0PJ]], $\\gamma'(t + c) = f(\\gamma(t + c)) = f(\\delta(t))$.\n7. Hence $\\delta'(t) = f(\\delta(t))$ for each $t \\in I - c$, and $\\delta$ is a solution of $x' = f(x)$ on $I - c$.\n\n## Idea\n\nThe right side $f(x)$ does not depend on time. Shifting the time variable does not change the derivative of a curve, so the shifted curve still has velocity $f$ at its own position.\n",
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