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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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In a symmetric 2-design, distinct blocks meet in exactly λ points

Statement

Let (P,B) be a symmetric 2-(v,k,λ) design. Then every two distinct blocks of B meet in exactly λ points.

Facts & Assumptions

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

[L1]

The counting identities are bk=vr and r(k1)=λ(v1) (A 2-design satisfies bk=vr and r(k1)=λ(v1)).

[L2]

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

Proof

technique · direct
1.1

Symmetry gives b=v, so the first identity in [L1] yields r=k.

L1algebra
2.1

Using step 1.1 in the second identity of [L1], one gets k(k1)=λ(v1), so kλ=k(vk)/(v1)>0 because k<v.

step 1.1L1algebra
3.1

If xTN=0, then xTNNTx=0, but [L2] and step 2.1 give xTNNTx=(kλ)ixi2+λ(ixi)2, which is positive for every nonzero x. Hence N is invertible.

step 2.1L2algebra
4.1

Every row and every column of N has sum k: rows because each point lies in r=k blocks by step 1.1, and columns because every block has size k. Therefore NJv=JvN=kJv, so step 3.1 gives N1JvN=Jv.

step 1.1step 3.1L1algebra
5.1

Multiplying the identity of [L2] on the left by N1 and on the right by N gives NTN=(kλ)Iv+λJv.

step 1.1step 4.1L2algebra
6.1

The (B,C) entry of NTN counts the points in BC, so step 5.1 shows that every off-diagonal entry is λ. Thus distinct blocks meet in exactly λ points.

step 5.1algebra

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Dependency tree · two levels

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