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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Fisher's inequality: every 2-(v,k,λ) design has at least v blocks

Statement

Every 2-(v,k,λ) design has at least v blocks.

Facts & Assumptions

Given: A 2-(v,k,λ) design with incidence matrix N and b blocks.

[L1]

The incidence identity is NNT=(rλ)Iv+λJv (For a 2-design, NNT=(rλ)I+λJ).

[L2]

The counting identities give r(k1)=λ(v1), with 2k<v and λ1 (A 2-design satisfies bk=vr and r(k1)=λ(v1)).

Proof

technique · direct
1.1

From [L2] one gets rλ=λ(vk)/(k1)>0, because λ1 and v>k.

L2algebra
2.1

If xRv is nonzero, then [L1] gives xTNNTx=(rλ)ixi2+λ(ixi)2>0 by step 1.1. Therefore no nonzero vector satisfies xTN=0.

step 1.1L1algebra
3.1

So the v rows of N are linearly independent in Rb. A family of v linearly independent vectors in Rb requires vb.

step 2.1algebra

Remarks

The positivity argument is over R. The published false statement FALSE: distinct nonempty A1,,Am[n] whose pairwise intersections all have the same parity satisfy mn records why the same proof does not survive over F2.

Depends on

Used by

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources