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`If all appearances of n numbers (2>1) together fill a space that defines a bug (appear 2 times in each unit they occupy), then the remaining appearances of these numbers will also form a bug.`

[edit: Changed 'n' to '2' as there is no need to consider larger groups.]

And here's the proof:

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`In any sudoku all numbers appear once in each row, column and box. This means that any group G of 2 different numbers fills 2 spaces in each of the 9 rows, columns and boxes. Removing any amount k of groups G where all the instances can be defined by k columns, rows and boxes, will leave us with 9-k columns, rows and boxes, where each unit will have exactly 2 spaces reserved for the 2 numbers of group G.`

That was complicated, an example to show what I mean:

The least amount of data we need to make a reduction based on uniqueness is this:

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`1..|...|...`

...|...|...

2..|1..|...

-----------

...|...|...

...|...|...

...|...|...

-----------

...|...|...

...|...|...

...|...|...

Assume these are the only given, or logically solved, instances of numbers 1 and 2 on the grid. If we placed number 2 in r1c4, then we would have two numberpairs that can be defined by two rows (1 and3) two columns (1 and 4) and 2 boxes (1 and 2). That means that in each other unit left on the grid there has to be exactly two spaces reserved for numbers 1 and 2, which in the end will lead to a bug => we can safely remove candidate 2 from r1c4.

Do you aprove? Object? Know this already?

-RW