1) First eliminations : e5=1%block, a7=8%block, h8=9%col, g9=7%block, i9=1%block, e9=2%row, h6=7%row, i2=5%cell, h4=1%col lead to 35 filled cells.
2) Now :
Look at only possibles g7=6,i7=6 in their block. They forbid{d7=6, b7=6, e7=6}.
Look at only possibles b7=9,c9=9 in their block. Whether c9=9 (in which case f2=9%col) or b7=9, in both cases, we have no more {b2=9}.
Look at only possibles i3=2,h3=2 in their row. Whether i3=2 (in which case f3=4%row) or h3=2 (in which case f3=8%row), in both cases, we have no more {f3=6, f3=5, f3=3}.
3) Look at e2. Whether e2=3, e2=7 (in which case d2=3%cell), or e2=9 (in which case e7=3%cell), we don't have e3=3. Now easy fillings up to 81 filled cells. (If needed, c3=3%row, b2=1%cell, g1=1%row, d1=4%row, f8=4%col, b7=4%block, a6=4%block, c6=1%block, i3=4%col, h3=2%block, h5=8%col, c4=8%row, c5=2%block, d4=2%block, f3=8%row, a8=1%block, g2=8%col, g5=4%block, a4=5%block, g6=5%col, f5=5%col, d8=5%col, b4=9%row, c1=5%block, a1=9%block, a3=6%block, c9=9%col, c8=6%col, e1=6%col, f9=6%block, f6=3%cell, d5=6%block, b5=3%cell, d7=3%cell, e2=3%block, f2=9%block, d2=7%row, e8=7%col, g7=2%col, i7=6%block, i4=3%cell, g4=6%col, b6=6%row, i6=2%col, e7=9%block, e3=5%cell)
Short hints for a proof
Beware, non-fc step needed here:
1) easy from 26 to 35 filled. 2) Looking at 6s in Bh8, eliminate some possibles. 3) Looking at 9s in Bb8, eliminate b2=9. 4) Looking at 2s in R3, eliminate f3=356. 5) Looking at e2, eliminate e3=3. Then it's easy fillings to end.
Forbidding-chain-like proof
Non-fc step is needed here :
around 35 filled
(g7=6)==(i7=6) forbids {d7=6, e7=6, b7=6}
(b7=9)==(c9=9)--(f9=9)==(f2=9) forbids {b2=9}
(f3=4)==(i3=4)--(i3=2)==(h3=2)--(h3=8)==(f3=8) forbids {f3=6, f3=3, f3=5}
(d2=3, d2=7)--(e2=7, e2=3, e2=9)--(e7=9, e7=3) gives that one of d2,e2,e7 is 3, which forbids e3=3. This one is no more a pure fc, so try to understand the way it works, I'll soon explain this generalization of fcs.
For advanced readers : One-elimination proof
After easy fillings to 35 filled, look only at these thick sets :
S1 : {f3=4, d1=4}: 4s in Be2 S2 : {d1=4, g1=4} : 4s in R1 S3 : {e7=3, d7=3} : 3s in R7 S4 : {e3=3, f3=3, c3=3} : 3s in R3 S5 : {g1=1, g2=1} : 1s in Bh2 S6 : {b7=9, e7=9} : 9s in R7 S7 : {b2=3, b2=1, b2=9} : 9s in Cb.
If b2=9, e7=9 (S6) ; if b2=3, e3 or f3=3 (S4) ; if b2=1, g1=1 (S5), d1=4 (S2) ; so (S7) one of e7=9, e3=3, f3=3,d1=4 is true. Now if e7=9 or e3=3, d7=3 (S3) ; and if f3=3, d1=4 (S1). So one of d7=3, d1=4 is true, which forbids d7=4.
Now easy fillings to 81 filled : b7=4%row, c9=9%block, e7=9%block, f2=9%col, a1=9%block, b4=9%block, f3=8%col, h5=8%col, g2=8%col, c4=8%block, a6=4%block, g5=4%block, i3=4%block, d1=4%block, f8=4%block, h3=2%block, g1=1%col, d7=3%block, f9=6%cell, a8=1%col, c8=6%block, a3=6%block, e1=6%block, e3=5%block, d8=5%block, e8=7%block, d2=7%block, c1=5%block, a4=5%block, g6=5%block, f5=5%block, f6=3%block, i4=3%col, e2=3%block, b5=3%col, b6=6%block, g4=6%block, i7=6%block, d5=6%block, i6=2%block, d4=2%block, c5=2%block, g7=2%col, c3=3%block, c6=1%col, b2=1%col.
How do you feel in front of such a proof ? (As for myself I would keep them for supertough puzzles when nothing simpler seems to work. However, it only needs to look inside 7 thick sets while the above proof needs to check 9 sets. That's why, to estimate the weigh of a proof I think necessary to consider the depth as well as the total number of steps)