05/12/27 tough puzzle from sudoku.com.au

A complete proof

Filled cells are left blank on diagrams for easier reading.
First eliminations : [b2=7%col, a5=7%col, h8=7%col] lead to 23 filled cells.

Look at only possibles b7=8,b9=8 in their col. They forbid{a7=8, c8=8, a8=8, c7=8}.
Look at only possibles h3=2,h1=2 in their col. They forbid{i3=2, i2=2, g3=2, g2=2}.
Now easy fillings up to 46 filled cells. (If needed, d8=8%row, f2=2%row, e3=8%col, a2=8%row, e7=2%col, e6=7%col, f6=1%row, c6=3%row, c5=8%block, i8=2%block, a8=1%row, g5=2%row, c8=4%block, f8=5%cell, a1=4%row, a4=5%cell, c7=5%block, c4=2%cell, h3=2%row, b1=2%row, b4=1%cell, a3=3%block, a7=9%col)

Look at only possibles c1=6,c1=9 in their cell. Whether c1=9 (in which case h1=6%cell) or c1=6, in both cases, we have no more {d1=6, e1=6, f1=6}.
Look at only possibles c1=9,c1=6 in their cell. Whether c1=6 (in which case h1=9%cell) or c1=9, in both cases, we have no more {d1=9, f1=9, e1=9}.
Now easy fillings up to 51 filled cells. (If needed, f1=3%cell, d5=3%row, e5=5%block, d1=5%row, e1=1%row)

Look at only possibles f9=9,f3=9 in their col. Whether f9=9 (in which case e9=6%cell) or f3=9 (in which case d2=6%cell), in both cases, we have no more {d9=6, d7=6}.
Look at only possibles f9=9,f3=9 in their col. Whether f9=9 (in which case e9=6%cell) or f3=9 (in which case f7=7%col), in both cases, we have no more {f7=6}.
Now easy fillings up to 54 filled cells. (If needed, f7=7%cell, d3=7%block, d7=1%cell)

Now, Look at only possibles e9=6,f9=6 in their block. They forbid{g9=6, h9=6, i9=6}.
Now easy fillings up to 81 filled cells. (If needed, h9=8%cell, b9=3%cell, h6=9%cell, i6=5%cell, g6=8%cell, c1=9%row, b7=8%row, c3=6%cell, g9=5%col, i9=1%row, g3=1%row, g4=4%col, d9=4%col, i3=4%col, f5=4%col, f3=9%row, i2=9%row, i7=3%col, g7=6%block, g2=3%col, e9=9%row, d4=9%row, d2=6%cell, f9=6%col, h1=6%cell, i5=6%cell, e4=6%block)

Short hints for a proof

One of the two heptagons is highlighted.

1) First eliminations to 23 filled cells.
2) Look at only possibles 8 in Cb. Look at only possibles 2 in Ch. Remove some possibles. Now easy eliminations take place again, leading to 46 filled cells.
3) Look at only possibles 69 in c1h1. Remove some possibles. Then simple eliminations take place again, up to 51 filled cells.
4) Looking at only possibles 9 in Cf, remove d9=6,d7=6,f7=6 (beware, heptagons). Now easy fillings up to 54 filled cells.
5) Look then at only possibles 6 in Be8. Remove some possibles. From here, simple eliminations lead to unique solution.

Forbidding-chain-like proof

One of the two heptagons is highlighted.

around 23 filled
(b7=8)==(b9=8) forbids {c7=8, a7=8, a8=8, c8=8}
(h3=2)==(h1=2) forbids {i3=2, g3=2, g2=2, i2=2}
around 46 filled
(c1=6)==(c1=9)--(h1=9)==(h1=6) forbids {d1=6, e1=6, f1=6}
around 51 filled
(e9=6)==(e9=9)--(f9=9)==(f3=9)--(d2=9)==(d2=6) forbids {d9=6, d7=6}
(e9=6)==(e9=9)--(f9=9)==(f3=9)--(f3=7)==(f7=7) forbids {f7=6}
around 54 filled
(e9=6)==(f9=6) forbids {i9=6, g9=6, h9=6}

Equivalence with 09/24

Now applying the same "dictionary" (renaming numbers, replacing cells) to proof of 05/09/24 you will mechanically get a proof of today's. E.g. first filled cell in 05/09/24 was i7=1%col; in today's it becomes b2=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 05/09/24's proof).