05/12/28 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : h7=6%row, b6=6%row, i4=6%row, e7=4%row, i8=4%row, c3=6%row lead to 27 filled cells.

2) Now :
Look at only possibles f6=7,d6=7 in their row. They forbid{d5=7, f5=7}.
Look at only possibles i9=2,h9=2 in their block. They forbid{a9=2, f9=2, b9=2, e9=2}.
Look at only possibles b7=7,c7=7 in their row. They forbid{b9=7}.
Look at only possibles d9=6,f9=6 in their row. Whether f9=6 (in which case d9=7%block) or d9=6, in both cases, we have no more {d9=1, d9=5}.
Look at only possibles f9=6,d9=6 in their row. Whether d9=6 (in which case f9=7%block) or f9=6, in both cases, we have no more {f9=2, f9=1}.
Now easy fillings up to 29 filled cells. (If needed, e4=2%col, f8=2%col)

3) Look at only possibles f4=8,f5=8 in their block. They forbid{f1=8, f2=8}.
Look at only possibles b3=1,e3=1 in their row. Whether e3=1 (in which case d8=1%block) or b3=1, in both cases, we have no more {b8=1}.

4) Here is the hardest step, so as suggested by _unknown, I highlighted the vertices of the 11-agon involved !
Look at only possibles c5=9,c7=9 in their col. Whether c5=9 (in which case b4=8%cell,b8=5%cell) or c7=9 (in which case g7=1%cell,d8=1%row), in both cases, we have no more {d8=5}.
Now easy fillings up to 50 filled cells. (If needed, e9=5%block, e6=3%cell, h4=3%row, a6=4%cell, d1=3%col, b8=5%row, d8=1%cell, d4=9%cell, f6=7%cell, f9=6%cell, d6=5%cell, h6=9%cell, d9=7%cell, f4=1%row, f5=8%col, d5=4%row, d2=6%cell, i5=7%cell, i3=3%cell, g3=7%block, b4=8%cell)

5)Look at only possibles f1=9,f1=4 in their cell. Whether f1=9 (in which case i1=2%cell) or f1=4 (in which case h2=4%row), in both cases, we have no more {h2=2}.
Look at only possibles g7=9,g7=1 in their cell. Whether g7=9 (in which case c5=9%col) or g7=1 (in which case a7=2%cell,a5=3%cell), in both cases, we have no more {c5=3}.
Look at only possibles i9=9,i1=9 in their col. Whether i9=9 (in which case b9=1%cell) or i1=9 (in which case f1=4%cell,b3=4%col), in both cases, we have no more {b3=1}.
Now easy fillings up to 81 filled cells. (If needed, b3=4%cell, a5=3%block, g9=3%row, c8=3%row, g8=8%row, e3=1%row, e1=8%block, h3=8%row, c2=8%col, a2=2%row, a1=5%block, a7=1%cell, b9=9%cell, c5=9%col, g7=9%block, f2=9%row, f1=4%block, h2=4%row, g2=5%block, h5=5%block, g5=1%row, h9=1%row, c1=7%cell, b1=1%cell, b7=7%col, h1=2%cell, i1=9%cell, a9=8%col, i9=2%row, c7=2%col, b5=2%col)

Short hints for a proof

The crux is at step 4 :

1) easy to 27 filled.
2) Looking at 7 in R7, at 2 in block Bh8, at 7 in R6, at 67 in d9f9, eliminate some possibles. Then easy to 29 filled.
3) Looking at 8 in Be5, eliminate some possibles. Looking at 1 in R3, eliminate b8=1 (beware, pentagon).
4) Looking at 9 in Cc, eliminate d8=5 (beware, 11-agon ! but what a reward to get it by yourself...). Then easy to 50 filled.
5) Looking at f1, eliminate h2=2 (heptagon). Looking at g7, eliminate c5=3 (nonagon). Looking at 9s in Ci, eliminate b3=1(nonagon).Now easy to unique solution.

Forbidding-chain-like proof

The 11-agon is needed here :

around 27 filled
(f6=7)==(d6=7) forbids {d5=7, f5=7}
(i9=2)==(h9=2) forbids {e9=2, a9=2, f9=2, b9=2}
(b7=7)==(c7=7) forbids {b9=7}
(d9=6)==(f9=6)--(f9=7)==(d9=7) forbids {d9=1, d9=5}
(f9=6)==(d9=6)--(d9=7)==(f9=7) forbids {f9=1, f9=2}
around 29 filled
(f4=8)==(f5=8) forbids {f1=8, f2=8}
(b3=1)==(e3=1)--(e9=1)==(d8=1) forbids {b8=1}
(b8=5)==(b8=8)--(b4=8)==(b4=9)--(c5=9)==(c7=9)--(g7=9)==(g7=1)--(g8=1)==(d8=1) forbids {d8=5}
around 50 filled
(i1=2)==(i1=9)--(f1=9)==(f1=4)--(f2=4)==(h2=4) forbids {h2=2}
(c5=9)==(c7=9)--(g7=9)==(g7=1)--(a7=1)==(a7=2)--(a5=2)==(a5=3) forbids {c5=3}
(b9=1)==(b9=9)--(i9=9)==(i1=9)--(f1=9)==(f1=4)--(b1=4)==(b3=4) forbids {b3=1}

Equivalence with 09/20

Now applying the same "dictionary" (renaming numbers, replacing cells) to proof of 05/09/20 you will mechanically get a proof of today's. E.g. first filled cell in 05/09/20 was a7=9%row; in today's it becomes h7=6%row. Try it ! Or see the method explained step by step on 10/08 example.
In case that you be lazy enough not to find it by yourself, see hints for today's proof (automatically derived from 05/09/20's proof).