1:
15(5) = {12345}
D\,D/: 20(5),21(5) each must have exactly one set of {67}
--> 20(5) = {12467}, 21(5) = {12567/13467}
2:
D\ + D/ = 28 + 28 = 56 = 15(5) + 20(5) + 21(5)
--> R4C4 = R15C4
NC: R145C4 = [231/253/451] (NC: R5C35 <> [2])
--> [1] of 20(5) locked in D\12+D/12
--> 15(5),20(5) each has one [1] of D\,D/
--> D\67+D/67 <> [1]
21(5): R5C4 = [1]
3:
Hidden pair 15(5): R3C35 = {12} (R3,R24C35)
15(5): D\45 & D/45 must each have two of {345}
NC: R4C35 <> {345}, = {67} (R4)
NC: R14C4 = [23] (step 2)
--> R5C35 = {45} (R5,R6C35)
21(5): D\67+D/67 = {2567} (step 1)
4:
20(5): D\12+D/12 = {1467} (step 1)
[2] of R2 locked in R2C17
NC: R2C17 <> {12} ([1] in D\12+D/12), <> [1]
--> [1] of R2 locked in R2C26 (R1C17)
--> [1] of C7 locked in R46C7 (NC: R5C7 <> [2])
--> 10(3) = [316/613] (C7)
5:
Hidden single C1: R6C1 = [1]
--> [3] of R6 locked in R6C35 (NC: R6C4 <> [4])
Hidden single R6: R6C7 = [4]
NC: C7 = [7531642]
Singles and simple NC placements from here.
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