1. The solver stops here:
AC7 is [234] or [345] eliminating 4 from FJ7.
Every row must have a 5, so every column must also have a 5 (Setti), solving J9=5
(the only 5 in column 9).
Every column except 9 must have a 4, and column 9 cannot have a 4. Every row
except perhaps row E must have a 4, so row E cannot have a 4 (Setti). Thus 4 and
stranded lower numbers can be eliminated from E15, solving E2 for 7 which
eliminates the 1s from EJ2 and solves F2=2.
If F8 were 6, EF23 would be a prohibited formation (assuming the uniqueness of the
puzzle) solving F8=9 and eliminating 5 from G8. With 5 eliminated from FJ8, since
every column must have a 5 (deduced above), AB8 must contain a 5 and therefore
cannot contain a 3.
EF34 is an x-wing on 8. eliminating all other 8s in columns 3 and 4.
GJ56 is an x-wing on 2, eliminating all other 2s in column 5 and eliminating the
stranded 1 in D5.
Every row must have an 8 so every column must also have an 8, solving A9=8 (the
only 8 in column 9) and thus A6=7. The resultant naked pair 56 in AD6 solves D6=8
(9 would be out of sequence with the necessary 5), and eliminating the 8 from D5
reveals the naked quad 1379 in column 5, eliminating those numbers from G7 and
J7.
Counting 2s, five columns must have a 2 and three columns cannot have a 2. It is
unknown whether column 7 contains a 2. Five rows must have a 2, two rows cannot
have a 2, and it is unknown whether row B and whether row C has a 2. By Setti, one
of those rows must have a 2, so AC7 must contain a 2 (the only 2s in rows B and C
are in column 7). Thus 5 can be eliminated from AC7, thereby solving A7=4, A8=5,
B8=4, G7=5 (column 7 must have a 5 and G7 is the only remaining one), J7=8 and
H7=6.
B48 must be [12345], [23456] or [34567]. It cannot be [12345], because the only 1 is
B5 and the only 2 is B7, but those cells are also the only 3s, so 1 can be eliminated

2. from B5, solving C5=1 (the only 1 in AJ5) eliminating 67 from C37, solving C6=5,
C7=2, C4=4 and C3=3. The rest is simple routine leading to the solution: