bennys wrote:I think i can see another step but it doesn't help much

A can't be 4

because then C and D force B to be 2

So F is 4 and then you get that E is 8 but thats all.

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`+-------+-------+-------+`

| A B 1 | 9 . 7 | C D 8 |

| 6 . . | 1 8 5 | 7 3 . |

| . . 7 | 4 6 . | 1 . . |

+-------+-------+-------+

| . 3 4 | . 9 . | . . . |

| . . . | 5 . 4 | . . . |

| . . . | . 1 . | 4 2 . |

+-------+-------+-------+

| E . 5 | . 7 1 | 9 F . |

| . 1 . | 8 4 . | . . 7 |

| 7 . . | . 5 9 | 2 . . |

+-------+-------+-------+

Well done, Bennys. Actually, you are so close from solving it from here. All you need is an xyz-chain as follows:

r7c2=2 => r7c4=r9c4=36 (naked pair) => r8c6<>6

r7c2=6 => r7c9<>6 => r8c7=6 or r8c8=6 => r8c6<>6

The rest is trivial.

BTW, there are at least 4 double implication forcing chains to be identified from the original grid.

Chain 1 enables a 4 to be removed from r7c9 (turbot fish)

Chain 2 enables a 2 to be removed from r7c1

Chain 3 enables an 8 to be removed from r7c1

Chain 4 enables a 4 to be removed from r7c8

Chain 2 and chain 3 would have fixed a 4 in r7c1 and an 8 in r7c8 also.

Double implication forcing chain and xyz-chain are both non-T&E.