05/12/21 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : a3=3%row, i7=4%row, e3=8%row, e7=2%cell, a7=8%cell, g3=5%row, h3=4%row, g6=4%block, i3=7%cell lead to 28 filled cells.

2) Now :
Look at only possibles h6=7,h4=7 in their block. They forbid{h8=7, h9=7}.
Look at only possibles g2=6,g2=1 in their cell. Whether g2=1 (in which case g1=6%cell) or g2=6, in both cases, we have no more {g9=6, h1=6, i2=6, g5=6, g8=6}.
Look at only possibles g2=1,g2=6 in their cell. Whether g2=6 (in which case g1=1%cell) or g2=1, in both cases, we have no more {g8=1, g9=1, i2=1, g5=1, h1=1}.
Look at only possibles b7=3,b7=7 in their cell. Whether b7=7 (in which case c7=3%cell) or b7=3, in both cases, we have no more {c9=3, b9=3}.
Look at only possibles b7=7,b7=3 in their cell. Whether b7=3 (in which case c7=7%cell) or b7=7, in both cases, we have no more {b9=7, c9=7, c8=7}.

3) Now :
Look at only possibles c9=2,c9=1 in their cell. Whether c9=1 (in which case c8=2%cell) or c9=2, in both cases, we have no more {b9=2, c5=2, c1=2, a8=2, c2=2}.
Look at only possibles c9=1,c9=2 in their cell. Whether c9=2 (in which case c8=1%cell) or c9=1, in both cases, we have no more {a8=1, c5=1, c4=1}.
Look at only possibles g5=8,g5=3 in their cell. Whether g5=3 (in which case c5=8%cell) or g5=8, in both cases, we have no more {h5=8, b5=8, f5=8, d5=8}.
Look at only possibles g5=3,g5=8 in their cell. Whether g5=8 (in which case c5=3%cell) or g5=3, in both cases, we have no more {f5=3, d5=3, b5=3}.

4) Look at only possibles e9=9,e4=9 in their col. Whether e9=9 (in which case b9=6%cell) or e4=9 (in which case f5=6%cell), in both cases, we have no more {b5=6}.
Now easy fillings up to 41 filled cells. (If needed, b5=2%cell, d6=2%col, d5=1%cell, e1=1%col, g2=1%col, g1=6%block, f2=6%block, f5=9%cell, e9=9%block, a8=9%block, b9=6%block, h5=6%cell, i8=6%col)

5)Look at only possibles b6=8,b6=3 in their cell. Whether b6=3 (in which case c5=8%cell) or b6=8, in both cases, we have no more {b4=8, c4=8}.
Look at only possibles b6=3,b6=8 in their cell. Whether b6=8 (in which case c5=3%cell) or b6=3, in both cases, we have no more {c4=3, b4=3}.
Now easy fillings up to 47 filled cells. (If needed, b4=4%cell, c4=5%cell, e6=5%col, e4=6%block, a6=6%block, a4=1%col)

6)Look at only possibles c5=3,g5=3 in their row. Whether c5=3 (in which case b6=8%cell) or g5=3 (in which case i4=9%cell,b2=9%row), in both cases, we have no more {b2=8}.
Now easy fillings up to 81 filled cells. (If needed, b2=9%cell, h1=9%block, i4=9%block, i2=2%cell, a2=5%cell, d2=4%cell, c1=4%row, f8=4%row, c7=7%col, c5=3%col, b7=3%col, b6=8%cell, g5=8%cell, g8=7%cell, g9=3%cell, b1=7%col, c2=8%cell, a1=2%cell, i6=3%col, h6=1%block, i9=1%block, c8=1%col, h8=2%row, c9=2%row, h9=8%cell, d8=8%block, f4=8%row, d4=3%row, f1=3%row, d9=7%col, f9=5%cell, d1=5%cell, h4=7%block, f6=7%block)

Short hints for a proof

Beware, 9-agon needed here:

1) easy to 28 filled.
2) looking at 7 in Bh5, at 37 in b7c7, at 16 in g1g2, eliminate some possibles.
3) looking at 12 in c8c9, at 38 in g5c5, eliminate some possibles.
4) looking at 9 in Ce, eliminate b5=6 (beware, heptagon). Now easy to 41 filled.
5)looking at 38 in b6c5, eliminate some possibles. Now easy to 47 filled.
6) looking at 3 in R5, eliminate b2=8 (beware, nonagon). Then easy to unique solution.

Forbidding-chain-like proof

The 9-agon is needed here :

Around 28 filled :
(h6=7)==(h4=7) forbids {h9=7, h8=7}
(g2=6)==(g2=1)--(g1=1)==(g1=6) forbids {g9=6, h1=6, i2=6, g8=6, g5=6}
(g2=1)==(g2=6)--(g1=6)==(g1=1) forbids {g8=1, h1=1, g5=1, g9=1, i2=1}
(b7=3)==(b7=7)--(c7=7)==(c7=3) forbids {c9=3, b9=3}
(b7=7)==(b7=3)--(c7=3)==(c7=7) forbids {b9=7, c9=7, c8=7}
(c9=2)==(c9=1)--(c8=1)==(c8=2) forbids {b9=2, a8=2, c1=2, c2=2, c5=2}
(c9=1)==(c9=2)--(c8=2)==(c8=1) forbids {a8=1, c4=1, c5=1}
(g5=8)==(g5=3)--(c5=3)==(c5=8) forbids {d5=8, f5=8, b5=8, h5=8}
(g5=3)==(g5=8)--(c5=8)==(c5=3) forbids {f5=3, d5=3, b5=3}
(b9=6)==(b9=9)--(e9=9)==(e4=9)--(f5=9)==(f5=6) forbids {b5=6}
Around 41 filled :
(b6=8)==(b6=3)--(c5=3)==(c5=8) forbids {c4=8, b4=8}
(b6=3)==(b6=8)--(c5=8)==(c5=3) forbids {c4=3, b4=3}
Around 47 filled :
(b6=8)==(b6=3)--(c5=3)==(g5=3)--(i4=3)==(i4=9)--(i2=9)==(b2=9) forbids {b2=8}

Equivalence with 09/12

Now applying the same "dictionary" (renaming numbers, replacing cells) to proof of 05/09/12 you will mechanically get a proof of today's. E.g. first filled cell in 05/09/12 was c7=8%col; in today's it becomes a3=3%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 05/09/12's proof).