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Equation 1 · What a Proof Assistant Changes

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∀V. packing⁡(V)  ⟹  ∃c. ∀r. 1≤r  ⟹  card⁡ ⁣(V∩B(0,r))≤πr318+c r2\forall V.\ \operatorname{packing}(V) \implies \exists c.\ \forall r.\ 1 \le r \implies \operatorname{card}\!\left(V \cap B(0,r)\right) \le \frac{\pi r^{3}}{\sqrt{18}} + c\,r^{2}

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This equation states a bound: one expression must stay on the indicated side of the other under the article’s assumptions. Read the equation part by part below; each part has a contextual explanation and a link to its mathematical background.

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VV

Symbol V

V is a part of this expression. Its role is fixed by the surrounding article and by the operations shown in the formula.

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cc

Symbol c

c is a part of this expression. Its role is fixed by the surrounding article and by the operations shown in the formula.

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rr

Symbol r

r is a part of this expression. Its role is fixed by the surrounding article and by the operations shown in the formula.

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BB

Symbol B

B is a part of this expression. Its role is fixed by the surrounding article and by the operations shown in the formula.

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π\pi

Symbol pi

pi occurs above the fraction bar. The numerator is divided by the entire denominator below it.

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r3r^{3}

Symbol r^3

r3r^3 occurs above the fraction bar. The numerator is divided by the entire denominator below it.

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r2r^{2}

Symbol r^2

The square of r: multiply r by itself.

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fraction

fraction

Divide the expression above the line by the one below it.

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√

√

Take a square root.

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addition

addition

Add the term after the plus sign to the term or group before it.

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superscript

superscript

A raised number can be a power. When it is a label or bound, it selects a case or the upper limit of a sum; the formula’s structure distinguishes these uses.

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πr3\pi r^{3}

Numerator: pi r^3

The complete quantity above the fraction bar.

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18\sqrt{18}

Denominator: sqrt18

The complete quantity below the fraction bar; it must be nonzero for this division.

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How to interpret it

With a fixed numerator, increasing a nonzero denominator reduces the fraction.

What the article says around this equation

Hales made the same move on the Kepler conjecture and describes the choices involved. The formal statement was deliberately engineered for readability: all mention of measure was removed, and the theorem was formulated as a claim about finite packings inside a large ball, which introduces a boundary error term but avoids importing measure theory into the statement. Cleaned up, it reads: ∀V. packing⁡(V)  ⟹  ∃c. ∀r. 1≤r  ⟹  card⁡ ⁣(V∩B(0,r))≤πr318+c r2\forall V.\ \operatorname{packing}(V) \implies \exists c.\ \forall r.\ 1 \le r \implies \operatorname{card}\!\left(V \cap B(0,r)\right) \le \frac{\pi r^{3}}{\sqrt{18}} + c\,r^{2}. Every symbol here is an invitation to a fidelity question. What does packing unfold to? Is the error term doing more work than it appears? Hales notes that the constant is not explicit in the statement, and that the theorem includes no uniqueness claim [ 4 ] . None of this is concealed —…
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Hales made the same move on the Kepler conjecture and describes the choices involved. The formal statement was deliberately engineered for readability: all mention of measure was removed, and the theorem was formulated as a claim about finite packings inside a large ball, which introduces a boundary error term but avoids importing measure theory into the statement. Cleaned up, it reads: ∀V. packing⁡(V)  ⟹  ∃c. ∀r. 1≤r  ⟹  card⁡ ⁣(V∩B(0,r))≤πr318+c r2\forall V.\ \operatorname{packing}(V) \implies \exists c.\ \forall r.\ 1 \le r \implies \operatorname{card}\!\left(V \cap B(0,r)\right) \le \frac{\pi r^{3}}{\sqrt{18}} + c\,r^{2}. Every symbol here is an invitation to a fidelity question. What does packing unfold to? Is the error term doing more work than it appears? Hales notes that the constant is not explicit in the statement, and that the theorem includes no uniqueness claim [ 4 ] . None of this is concealed — but none of it is checked by the kernel either.

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