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. . 47 | 45 . . | . . .
. . 57 | 45 . . | . . .
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Basically, the 45 pair up there cant be r5c4 = 4, r6c4 = 5 due to the fact that there would end up being two 7's left in c3.
This is relatively simple considering the level of solving used here, but I've never seen this particular technique addressed directly before. If it doesn't exist (as far as you guys know, anyway ) I would love to give it a name.