Tough Sudoku for 12/August/2021

Basics to 28
1. XYZ -wing 347 [g39,h3] =>g2<> 4 if g2=5 =>b2=0 =>b2=7 solve
i6=5=e8 f9=3=e6=c5 c6=4 c1=2 d9=6=f4 d8=8=e4=f2 e2=7

correction ( b2=7) =>g2=7
Basics to 28
1. XYZ -wing 347 [g39,h3] =>g2<> 4 if g2=5 =>b2=0 => g2=7 solve
i6=5=e8 f9=3=e6=c5 c6=4 c1=2 d9=6=f4 d8=8=e4=f2 e2=7

1. Unique possibilities to 28. Using Unresolvable Rectangles:
2. e4<>3, or else 19UR@gi48.
3. g9<>7,or else i9=4,g4=34(to avoid 19UR@gi48),creating 34UR@gh34. UP30.
4. Whether e8=3,(g5=5 after e26=5),OR e6=3=c5,cai1=245;g5=5.UP81.

Basics to 28
1. Note external guardians to UR(34)gh34 are (4)g2, (4)g9: (4)g9 == (4)g2 - (4=37)gh3 => -7g9 ; UP = 30
2. Now using internal guardians to UR(19)gi48,
(5=24)i12 - (4)i4 == (3)g4 - (3*)g5 = [ (5=2)i1-c1=(2-3)c5=*f5-f9=(3-5)e8= {SS(5):e6=e2-g2=g5} ] => -5i6 ; UP = 81

Basic techniques to UP28.
5g2=47g29-(47=3)g3-(3=4)h3 => g2<>4
If g2=5 then 5=i6=e8=h9, 8=d8=h7=e2=b6=f4, 6=d4 => d9=0, so g2=7. UP81

HI Jyrki. In your last step, after g2=5, you make e2=8. Can't e2 also be 7?
Best Regards, Alfred.

Correct, Alfred. There is a problem. I first used that logic to deduce b2<>7, and then thought that it works the same way, but it does not. Bummer.