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Published equation contexts

∀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}

Why this formula appears here

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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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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How to interpret it

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

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Published contexts (1)

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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}

Equation 1 · Mathematics & Computation

What a Proof Assistant Changes

This equation states a bound: one expression must stay on the indicated side of the other under the article’s assumptions.

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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