{
    "messages": [
        {
            "role": "system",
            "content": "As a mathematician familiar with Isabelle, your task is to translate the natural language proof into an Isabelle language version."
        },
        {
            "role": "user",
            "content": "Natural language version:\n(*### Problem\n\nThe fifth and eighth terms of a geometric sequence of real numbers are $7!$ and $8!$ respectively. What is the first term? The final Answer is 315\n\n### Proof\nSo we know that the fifth and eighth terms of a geometric sequence are $7!$ and $8!$.\nThat means that our two equations are $ar^4=7!$ and $ar^7=8!$.\nNow if we divide these two equations, we get $\\frac{ar^4}{ar^7}=\\frac{7!}{8!}$.\nSimplifying, we get $r^3=\\frac{1}{8}$.\nThat means $r = 2$.\nNow we can just plug in $r=2$ and $n=4$ into the formula for the nth term. That gives us $ar^3=7!$. We can divide both sides of this equation by $2^3$ to get $ar^3=7!$.\n$a*2^4=7!$, so $a = \\frac{7!}{2^4}$.\nSo the first term is $\\boxed{840}$.# Answer840\nThat's a big number. Let's check it.*)\n\nIsabelle version:\ntheorem\nfixes a r :: real\nassumes h0 : \"a > 0\"\nand h1 : \"r \\<noteq> 0\"\nand h2 : \"a * r^4 = fact 7\"\nand h3 : \"a * r^7 = fact 8\"\nshows \"a = 315\"\n proof- "
        },
        {
            "role": "assistant",
            "content": "  (* So we know that the fifth and eighth terms of a geometric sequence are $7!$ and $8!$. *)\n  (* That means that our two equations are $ar^4=7!$ and $ar^7=8!$. *)\n  have eq1: \"a * r^4 = fact 7\" using h2 sledgehammer\n  have eq1: \"a * r^7 = fact 8\" using h3 sledgehammer\n  (* Now if we divide these two equations, we get $\\frac{ar^4}{ar^7}=\\frac{7!}{8!}$. *)\n  have eq2: \"a * r^4 / (a * r^7) = fact 7 / fact 8\" using h2 h3 sledgehammer\n  (* Simplifying, we get $r^3=\\frac{1}{8}$. *)\n  then have eq3: \"r^3 = 1/8\" sledgehammer\n  (* That means $r = 2$. *)\n  then have eq4: \"r = 2\" sledgehammer\n  (* Now we can just plug in $r=2$ and $n=4$ into the formula for the nth term. That gives us $ar^3=7!$. *)\n  have eq5: \"a * (2^3) = fact 7\" using h2 eq4 sledgehammer\n  (* We can divide both sides of this equation by $2^3$ to get $ar^3=7!$. *)\n  have eq6: \"a * r^3 = fact 7\" sledgehammer\n  (* $a*2^4=7!$, so $a = \\frac{7!}{2^4}$. *)\n  have eq7: \"a = fact 7 / 2^4\" using h2 sledgehammer\n  (* So the first term is $\\boxed{840}$. *)\n  then have \"a = 840\" sledgehammer"
        },
        {
            "role": "user",
            "content": "Natural language version:\n(*### Problem\n\nSimplify\n\\[\\tan x + 2 \\tan 2x + 4 \\tan 4x + 8 \\cot 8x.\\]The answer will be a trigonometric function of some simple function of $x,$ like \"$\\cos 2x$\" or \"$\\sin (x^3)$\". The final Answer is \\cot x\n\n ### Proof\nSo we want to simplify $\\tan x + 2 \\tan 2x + 4 \\tan 4x + 8 \\cot 8x$.\nAnd we should start by changing the $\\cot$ to a $\\tan$ since the other terms are all $\\tan$ terms.\nRight. So we want to simplify $\\tan x + 2 \\tan 2x + 4 \\tan 4x + 8 \\frac{1}{\\tan 8x}$.\nAnd we should be careful when we multiply by $\\frac{1}{\\tan 8x}$ since we don't want to change the value of the expression.\nSince we want to get the expression in terms of $\\tan x$, we should try to change the $\\cos 8x$ to an expression in terms of $\\tan x$.\nSo we want to simplify $\\tan x + 2 \\tan 2x + 4 \\tan 4x + 8 \\frac{1}{\\tan 8x} - \\cot x + \\cot x$.\nThat's right. So we want to simplify $\\tan x + 2 \\tan 2x + 4 \\tan 4x + 8 \\frac{\\cos 8x}{\\sin 8x} - \\cot x + \\cot x$.\nThat way we can see that $\\tan x - \\cot x = -2 \\cot^3 x$\nSo the expression simplifies to $-2 \\cot 2x + 2 \\tan 2x + 4 \\tan 4x + 8 \\cot 8x + \\cot x$.\nNow we can use the identity $\\cot a + \\tan a = 2 \\cot 2a$ to simplify.\nAnd again to $-8 \\cot 8x + 8 \\tan 8x + \\cot x$.\nAnd now we can simplify to just $\\cot x$.\nRight, so $\\tan x + 2 \\tan 2x + 4 \\tan 4x + 8 \\cot 8x = \\cot x$.# Answer\\cot x*)\n\nIsabelle version:\ntheorem\nfixes x :: real and tan_x tan_2x tan_4x cot_8x :: real\nassumes h0 : \"tan_x = tan x\"\nand h1 : \"tan_2x = tan (2 * x)\"\nand h2 : \"tan_4x = tan (4 * x)\"\nand h3 : \"cot_8x = 1 / tan (8 * x)\"\nshows \"tan_x + 2 * tan_2x + 4 * tan_4x + 8 * cot_8x = 1 / tan x\"\n proof- "
        },
        {
            "role": "assistant",
            "content": "  (* So we want to simplify $\\tan x + 2 \\tan 2x + 4 \\tan 4x + 8 \\cot 8x$. *)\n  (* And we should start by changing the $\\cot$ to a $\\tan$ since the other terms are all $\\tan$ terms. *)\n  (* Right. So we want to simplify $\\tan x + 2 \\tan 2x + 4 \\tan 4x + 8 \\frac{1}{\\tan 8x}$. *)\n  have eq1: \"tan_x + 2 * tan_2x + 4 * tan_4x + 8 * cot_8x = tan x + 2 * tan (2 * x) + 4 * tan (4 * x) + 8 * (1 / tan (8 * x))\" using h0 h1 h2 h3 sledgehammer\n  (* And we should be careful when we multiply by $\\frac{1}{\\tan 8x}$ since we don't want to change the value of the expression. *)\n  (* Since we want to get the expression in terms of $\\tan x$, we should try to change the $\\cos 8x$ to an expression in terms of $\\tan x$. *)\n  (* So we want to simplify $\\tan x + 2 \\tan 2x + 4 \\tan 4x + 8 \\frac{1}{\\tan 8x} - \\cot x + \\cot x$. *)\n  (* That's right. So we want to simplify $\\tan x + 2 \\tan 2x + 4 \\tan 4x + 8 \\frac{\\cos 8x}{\\sin 8x} - \\cot x + \\cot x$. *)\n  (* That way we can see that $\\tan x - \\cot x = -2 \\cot^3 x$ *)\n  have eq2: \"tan_x - 1 / tan_x = -2 * (1 / tan_x)^3\" sledgehammer\n  (* So the expression simplifies to $-2 \\cot 2x + 2 \\tan 2x + 4 \\tan 4x + 8 \\cot 8x + \\cot x$. *)\n  have eq3: \"tan_x + 2 * tan_2x + 4 * tan_4x + 8 * cot_8x = -2 * 1 / tan_2x + 2 * tan_2x + 4 * tan_4x + 8 * cot_8x + 1 / tan_x\" sledgehammer\n  (* Now we can use the identity $\\cot a + \\tan a = 2 \\cot 2a$ to simplify. *)\n  have eq4: \"1 / tan_x + tan_x = 2 * 1 / tan_2x\" sledgehammer\n  (* And again to $-8 \\cot 8x + 8 \\tan 8x + \\cot x$. *)\n  have eq5: \"tan_x + 2 * tan_2x + 4 * tan_4x + 8 * cot_8x = -8 * cot_8x + 8 * tan 8x + 1 / tan_x\" sledgehammer\n  (* And now we can simplify to just $\\cot x$. *)\n  then have eq6: \"tan_x + 2 * tan_2x + 4 * tan_4x + 8 * cot_8x = 1 / tan_x\" sledgehammer\n  (* Right, so $\\tan x + 2 \\tan 2x + 4 \\tan 4x + 8 \\cot 8x = \\cot x$. *)\n  then then have eq7: \"tan_x + 2 * tan_2x + 4 * tan_4x + 8 * cot_8x = 1 / tan_x\" sledgehammer"
        }
    ]
}