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Equation 1 · Part 5 · What a Proof Assistant Changes

Symbol pi

∀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}
π\pi

What this part means

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

Its job in the formula

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

The passage around this formula

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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A variable is a named place for a value. Its letter is a local label: x can mean position in one formula and a data point in another.

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Sources cited in the surrounding passage

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