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The ring and then N5

After easy moves to set up the ring and n5 (using innies @ r5 & 10/3 @ /456):

Innies @ r5: r5c456 = 21 = {489|579|678}
10/3 @ /456 = {127|136|145|235}

=> 17/3 @ \456 = [746|458|269|962|476]

r2c3467 = [ 4957 | 7693 ]
r3467c2 = [ 9468 | 6795 ]
r3467c8 = [ 3976 | 7549 ]
r8c3467 = [ 6849 | 9576 ]

r46c2=[46|79] => \456 can't be [746]
r28c6+r4c2=[544|977] => r4c4+r5c456 can't be [4759|4957] => \456 can't be [458]

=> 17/3 @ \456 = [269|962|476] with 6 locked
=> 10/3 @ /456 = {127|136} with 1 locked
=> r5c456 = 21 = {579|678} with 7 locked
=> r46c46 = {2913|4612} with 2 locked

Innies @ c5: r159c5=17 => r19c5 can't have 5
13/3 @ r234c5: r4c5 from {34589} => r23c5 can't have 5
r28c6+r4c2=[544|977] => r4c4+r5c456 can't be [4579] => r5c456 can't be [579]

5 @ c5 locked @ r4678c5

r2c6+r8c46=[584|957] => exactly one of r2c6+r8c4 must be 5

Grouped turbot fish:
n8: either r78c5 or r8c4 must not be 5 => either r46c5 or r2c6 must be 5 => r5c6 can't be 5

=> r5c456=[768|867]

The rest is trivial stuff.

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1. 9 locked for R2 and N2 in R2C46
2. R46C5 = {45/38}: only options left with your 17(3), 10(3) and hidden 21(3) options, so R456C5 = {4[6]5} or {3[7]8}
3. R456C5 = [873] blocked by 15(3) R6C5: R78C5 can't equal 12, as 3, 7 and 8 are already used in C5.
4. R456C5 = [378] -> R5C4 = 9 -> R2C4 = 6: which leave no options for 13(3) at R2C5 as 6, is used in N2 and 7,8 in C5 and 9 was locked in step 1 and R23C5 can't equal 10 anymore.
5. This leaves: R456C5 = {4[6]5}

r8c46 = 12(2) r159c5 =17(3) r5c5 = 7 conflicts with C5 as if 7 on N5/R5 analysis r46c5 = {45}

With the Ring in place
R2c6r4c8 = {59} -> r2c8r4c6 <> 59
62c8r8c6 = {47} -> r6c6r8c8 <> 47

Innies @ r5: r5c456 = 21 = {489|579|678}
10/3 @ /456 = {127|136|145|235}

=> 17/3 @ \456 = [746|458|269|962|476]

r2c3467 = [ 4957 | 7693 ] =79..
r3467c2 = [ 9468 | 6795 ] =69..
r3467c8 = [ 3976 | 7549 ] =79..
r8c3467 = [ 6849 | 9576 ] =69.. r8c46 = 12(2)

r46c2=[46|79] => \456 can't be [746] and <> [647] r6c6<>7
r28c6+r4c2=[544|977] => r4c4+r5c456 can't be [4759|4957] => \456 can't be [458]
and <> [854] r6c6<>4

=> 17/3 @ \456 = [269|962|476] with 6 locked r5c5 = 6|7
=> 10/3 @ /456 = {127|136} with 1 locked
=> r5c456 = 21 = {579|678} with 7 locked
=> r46c46 = {2913|4612} with 2 locked
At this stage Matt has reduced to r5c5 to 6/7 without any hint of T&E (mine was sort of T&E)
innies on C5 -> r159c5 = 17/3
If r5c5 =6 r46c5 = {46} r5c46 = {78} r4c6r6c4 = {13} r4c4r6c6 = {29}
If r5c5 =7 r46c5 = {38} r5c46 = {59} r4c6r6c4 = {12} r4c4r6c6 = [46]
-> c5 <>38 r19c5 = {46}|[19] but r1c456 <>9 hence <> {179} -> r1c5<>1 -> {46} but r234c5 = {23}8 | {14}8 | {46}3 | {28}3 | {19}3 all these conflict with the {46} in r19 or the 3|8 in r6 so fail-> r5c5 =6
This is the same as Para’s approach so is it SCA or T&E?
Simple from here

Now that you've locked the 9 @ r2,n2 @ r2c46,
if r5c5=7 => r46c5={38}, r19c5=17-7=10 (=> 13/3 @ r234c5=[463|643|148|418])
But 9 @ n2 locked @ r2c46 so r1c5 can't be 9
Also 17/3 @ r1c456 can't be {179} so r1c5 can't be 1
=> r19c5={46} => 13/3 @ r234c5 must be from {12358}={238} => contradicts r46c5={38}
Therefore r5c5 can't be 7, must be 6.

Your "but r234c5 = {46}5" looks completely non-understandable to me.

8e) 14(3) <> 6 because {356} blocked by Killer pair (35) of 10(2)

8e) 14(3) <> 6 because {356} blocked by Killer pair (56) of Innies R5

8i) ! Consider combos of 10(3) -> R4C45 <> 9:
- i) 10(3) = {136} -> Innies R5 = {579} -> R4C45 <> 9
- ii) 10(3) = {145} -> R4C8 = 9 -> R4C45 <> 9

grouped xy-wing:

10(3) @ R5C789 has 5|6
R4C8 has 5|9
Innies R5 @ R5C456 has 6|9
=> R4C45 can't have 9

I think HATMAN would consider this "fishy".