Once you've determined the 1 in R7 has to be in C4 or C5, 3 is the only number left that could possibly be in R9C4. 1, 4, 5, 6, and 8 are already present in the box, 2 and 9 are already in the column, and 7 is already in the row. Even though the 3 isn't eliminated from R9C6, everything but the 3 is eliminated from R9C4.

Got it now! Thanks for explaining in so detail. That really really helps. I somehow missed that logic. I now recollect having solved puzzles using this technique. Maybe I am out of practice or just becoming non-intelligent. :D