05/12/15 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : d2=3%col, c1=3%row, f1=8%row, e1=7%row, c8=6%block lead to 26 filled cells.

2) Now :
Look at only possibles c9=9,b9=9 in their block. They forbid{e9=9, g9=9, h9=9}.
Look at only possibles a7=1,a8=1 in their block. They forbid{a6=1, a5=1, a3=1, a2=1}.
Look at only possibles f5=7,d5=7 in their block. They forbid{g5=7, c5=7, a5=7, i5=7}.
Look at only possibles f3=9,f3=5 in their cell. Whether f3=5 (in which case f2=9%cell) or f3=9, in both cases, we have no more {f8=9, e2=9, f7=9}.
Look at only possibles f3=5,f3=9 in their cell. Whether f3=9 (in which case f2=5%cell) or f3=5, in both cases, we have no more {f8=5, d3=5, e2=5, f7=5, f5=5}.
Look at only possibles i8=7,i4=7 in their col. Whether i4=7 (in which case i8=3%col) or i8=7, in both cases, we have no more {i8=4, i8=2}.
Look at only possibles i4=7,i8=7 in their col. Whether i8=7 (in which case i4=3%col) or i4=7, in both cases, we have no more {i4=2, i4=8, i4=9}.
Now easy fillings up to 39 filled cells. (If needed, i7=8%col, d9=8%col, g9=7%row, i4=7%col, i8=3%cell, a9=3%col, g4=3%col, g8=4%block)

3)
Look at only possibles h8=5,h8=2 in their cell. Whether h8=2 (in which case h9=5%cell) or h8=5, in both cases, we have no more {h1=5, g7=5, h7=5}.
Look at only possibles h8=2,h8=5 in their cell. Whether h8=5 (in which case h9=2%cell) or h8=2, in both cases, we have no more {h6=2, h4=2}.
Look at only possibles a7=5,a7=1 in their cell. Whether a7=1 (in which case a8=5%cell) or a7=5, in both cases, we have no more {c9=5, b9=5, a5=5, a2=5, a3=5}.
Now easy fillings up to 54 filled cells. (If needed, a5=2%cell, e4=2%row, f5=7%cell, d8=7%col, a8=1%row, d7=1%row, d5=5%cell, f8=2%col, h9=2%col, h8=5%col, a7=5%cell, e9=5%cell, a3=7%cell, c6=7%col, g6=2%row)

Next chain was found by mark from ny.
4) Here is the hardest step, so as suggested by _unknown, I highlighted the vertices of the heptagon involved !
Look at only possibles c5=1,c5=9 in their cell. Whether c5=1 (in which case b6=4%cell) or c5=9 (in which case c9=4%cell), in both cases, we have no more {b9=4}.
Now easy fillings up to 81 filled cells. (If needed, b9=9%cell, b4=5%cell, c9=4%cell, g1=5%row, g5=1%col, b6=1%row, c3=1%row, h1=1%row, h7=6%col, g2=6%col, i5=6%col, b1=6%col, h4=9%col, g7=9%col, c4=8%cell, b3=2%cell, i2=2%row, i3=9%col, f2=9%col, f3=5%col, a6=4%cell, c5=9%cell, h6=8%cell, a2=8%col, i1=4%col, b2=4%col, c2=5%row)

Short hints for a proof

Beware, 6-FC needed here:

1) easy to 26 filled.
2) Looking at 9 in Bb8, at 1 in Bb8, at 7 in Be5, at 37 in i4i8, at 59 in f2f3, eliminate some possibles, then easy to 39 filled.
3) Looking at 25 in h8h9, at 15 in a7a8, eliminate some possibles, then easy to 54 filled.
4) Looking at c5, eliminate b9=4 (beware, heptagon). Then easy to unique solution.

Forbidding-chain-like proof

The 6-FC is needed here :

around 26 filled
(c9=9)==(b9=9) forbids {g9=9, e9=9, h9=9}
(a7=1)==(a8=1) forbids {a2=1, a3=1, a5=1, a6=1}
(f5=7)==(d5=7) forbids {c5=7, i5=7, g5=7, a5=7}
(f3=9)==(f3=5)--(f2=5)==(f2=9) forbids {f8=9, f7=9, e2=9}
(f3=5)==(f3=9)--(f2=9)==(f2=5) forbids {f8=5, f7=5, f5=5, e2=5, d3=5}
(i8=7)==(i4=7)--(i4=3)==(i8=3) forbids {i8=9, i8=2, i8=4}
(i4=7)==(i8=7)--(i8=3)==(i4=3) forbids {i4=9, i4=2, i4=8}
around 39 filled
(h8=5)==(h8=2)--(h9=2)==(h9=5) forbids {g7=5, h1=5, h7=5}
(h8=2)==(h8=5)--(h9=5)==(h9=2) forbids {h6=2, h4=2}
(a7=5)==(a7=1)--(a8=1)==(a8=5) forbids {a5=5, b9=5, c9=5, a2=5, a3=5}
around 54 filled
(b6=4)==(b6=1)--(c5=1)==(c5=9)--(c9=9)==(c9=4) forbids {b9=4}

Equivalence with 2005/09/28

Now applying the same "dictionary" (renaming numbers, replacing cells) to proof of 05/09/28 you will mechanically get a proof of today's. E.g. first filled cell in 05/09/28 was f8=6%col; in today's it becomes d2=3%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 05/09/28's proof).