05/11/22 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : h1=8%col, b9=7%col, b5=5%col, b2=2%col, b3=8%col, f2=8%block, h5=1%cell, h9=5%cell, b1=9%cell lead to 28 filled cells.

2) Now :
Look at only possibles f3=9,d3=9 in their block. They forbid{g3=9, i3=9}.
Look at only possibles a2=4,a2=3 in their cell. Whether a2=3 (in which case c2=4%cell) or a2=4, in both cases, we have no more {a1=4, c3=4, e2=4, g2=4, i2=4}.
Look at only possibles a2=3,a2=4 in their cell. Whether a2=4 (in which case c2=3%cell) or a2=3, in both cases, we have no more {i2=3, a1=3, g2=3, e2=3, c3=3}.
Look at only possibles h7=7,h7=9 in their cell. Whether h7=9 (in which case h8=7%cell) or h7=7, in both cases, we have no more {g8=7, g7=7}.
Look at only possibles h7=9,h7=7 in their cell. Whether h7=7 (in which case h8=9%cell) or h7=9, in both cases, we have no more {i8=9, g8=9, g7=9}.

3) Now :
Look at only possibles g8=1,g8=3 in their cell. Whether g8=3 (in which case i8=1%cell) or g8=1, in both cases, we have no more {e8=1, i9=1, c8=1, a8=1, g7=1}.
Look at only possibles g8=3,g8=1 in their cell. Whether g8=1 (in which case i8=3%cell) or g8=3, in both cases, we have no more {d8=3, e8=3, i9=3}.
Look at only possibles e2=5,e2=7 in their cell. Whether e2=7 (in which case e8=5%cell) or e2=5, in both cases, we have no more {e4=5, e6=5, e3=5, e7=5}.
Look at only possibles e2=7,e2=5 in their cell. Whether e2=5 (in which case e8=7%cell) or e2=7, in both cases, we have no more {e4=7, e6=7, e7=7}.

4) Look at only possibles g5=6,d5=6 in their row. Whether g5=6 (in which case g7=4%cell) or d5=6 (in which case e6=4%cell), in both cases, we have no more {e7=4}.
Now easy fillings up to 41 filled cells. (If needed, e7=1%cell, f4=1%col, e4=3%cell, a2=3%col, c2=4%block, a6=4%block, e3=4%col, e6=6%col, g5=6%block, i9=6%block, g7=4%block, i1=4%block, c5=3%col)

5)Look at only possibles f7=5,f7=7 in their cell. Whether f7=7 (in which case e8=5%cell) or f7=5, in both cases, we have no more {d7=5, d8=5}.
Look at only possibles f7=7,f7=5 in their cell. Whether f7=5 (in which case e8=7%cell) or f7=7, in both cases, we have no more {d7=7, d8=7}.
Now easy fillings up to 47 filled cells. (If needed, d7=8%cell, d8=2%cell, d5=4%cell, f9=4%block, d9=3%cell, f5=2%cell)

6)Look at only possibles e8=7,e2=7 in their col. Whether e8=7 (in which case f7=5%cell) or e2=7 (in which case d1=6%cell,a7=6%col), in both cases, we have no more {a7=5}.
Now easy fillings up to 81 filled cells. (If needed, a8=5%block, c8=8%block, c7=9%block, h8=9%block, a4=8%block, i6=8%block, h7=7%block, e8=7%block, a7=6%block, c3=6%block, d1=6%block, f1=7%block, g2=7%block, i2=9%block, f7=5%block, e2=5%col, f3=3%col, d3=9%block, f6=9%block, g4=9%block, i8=3%col, g1=3%col, g6=2%col, i4=5%block, d6=5%block, d4=7%block, c6=7%block, g3=5%block, a9=2%col, c4=2%col, i3=1%col, g8=1%col, c9=1%col, a1=1%col)

Short hints for a proof

Beware, 9-agon needed here:

1) easy to 28 filled.
2) looking at 9 in Be2, at 34 in a2c2, at 79 in h7h8, eliminate some possibles.
3) looking at 13 in g8i8, at 57 in e2e8, eliminate some possibles.
4) looking at 6 in R5, eliminate e7=4 (beware, heptagon). Now easy to 41 filled.
5)looking at 57 in e8f7, eliminate some possibles. Now easy to 47 filled.
6) looking at 7 in Ce, eliminate a7=5 (beware, nonagon). Then easy to unique solution.

Forbidding-chain-like proof

The 9-agon is needed here :

Around 28 filled :
(f3=9)==(d3=9) forbids {i3=9, g3=9}
(a2=4)==(a2=3)--(c2=3)==(c2=4) forbids {a1=4, g2=4, i2=4, e2=4, c3=4}
(a2=3)==(a2=4)--(c2=4)==(c2=3) forbids {i2=3, a1=3, e2=3, c3=3, g2=3}
(h7=7)==(h7=9)--(h8=9)==(h8=7) forbids {g7=7, g8=7}
(h7=9)==(h7=7)--(h8=7)==(h8=9) forbids {g8=9, g7=9, i8=9}
(g8=1)==(g8=3)--(i8=3)==(i8=1) forbids {i9=1, a8=1, g7=1, e8=1, c8=1}
(g8=3)==(g8=1)--(i8=1)==(i8=3) forbids {e8=3, i9=3, d8=3}
(e2=5)==(e2=7)--(e8=7)==(e8=5) forbids {e6=5, e4=5, e3=5, e7=5}
(e2=7)==(e2=5)--(e8=5)==(e8=7) forbids {e7=7, e4=7, e6=7}
(g7=4)==(g7=6)--(g5=6)==(d5=6)--(e6=6)==(e6=4) forbids {e7=4}
Around 41 filled :
(f7=5)==(f7=7)--(e8=7)==(e8=5) forbids {d7=5, d8=5}
(f7=7)==(f7=5)--(e8=5)==(e8=7) forbids {d7=7, d8=7}
Around 47 filled :
(f7=5)==(f7=7)--(e8=7)==(e2=7)--(d1=7)==(d1=6)--(a1=6)==(a7=6) forbids {a7=5}

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 b9=7%col. 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).