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I finally decided to try some of the anti-backdoor pairs I found in my previous post.
To be clear about what I did:
1) the resolution path starts from the givens, with:
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singles ==> r9c6 = 1, r9c4 = 7, r9c5 = 5, r9c7 = 9, r9c9 = 8, r9c1 = 3, r6c3 = 3, r3c8 = 8, r2c9 = 3
Starting non trivial part of solution with the following RESOLUTION STATE:
56789 3 789 289 279 28 1 4 2567
4678 678 478 248 1 5 267 9 3
14579 157 2 6 3479 34 57 8 57
124678 1678 1478 5 2469 2468 2467 3 124679
12467 167 5 1249 23469 2346 8 167 124679
12468 9 3 1248 246 7 2456 156 12456
1578 1578 178 24 246 9 3 1567 14567
1579 2 179 3 8 46 4567 1567 14567
3 4 6 7 5 1 9 2 8
and these trivial Singles are not considered as steps.
2) Non counted steps between two eliminations can consist of Singles
3) Final steps (after the second whip[≥1] elimination is found) can only be Singles.
4) What I find is not only 2-step but also 2-elimination before stte.
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whip[7]: c4n1{r6 r5} - c4n9{r5 r1} - b2n8{r1c4 r1c6} - r1c3{n8 n7} - r2n7{c3 c7} - r3c9{n7 n5} - r3c7{n5 .} ==> r6c4 ≠ 8
singles ==> r4c6 = 8, r1c6 = 2, r2c7 = 2, r1c9 = 6, r1c1 = 5, r7c2 = 5, r2c2 = 8, r2c4 = 4, r2c3 = 7, r1c3 = 9, r1c4 = 8, r1c5 = 7, r8c3 = 1, r4c3 = 4, r7c3 = 8, r7c1 = 7, r8c1 = 9, r2c1 = 6, r3c2 = 1, r3c1 = 4, r3c6 = 3, r3c5 = 9, r7c4 = 2, r6c4 = 1, r5c4 = 9, r4c9 = 9, r4c1 = 1, r5c1 = 2, r6c1 = 8, r6c9 = 2, r4c5 = 2, r5c5 = 3
whip[2]: r7n4{c9 c5} - r6n4{c5 .} ==> r8c7 ≠ 4
stte
There are other possibilities, but all involve at least a whip[7].