y' R U R' U' R' F R2 U' R' U' R U R' F'
y x R U R' U' R' F R2 U' R' U' R U R' F'
y R U R' U' R' F R2 U' R' U' R U R' F'
x y' R U R' U' R' F R2 U' R' U' R U R' F'
y R U R' U' R' F R2 U' R' U' R U R' F'
x' z' R U R' U' R' F R2 U' R' U' R U R' F'
x' R U R' U' R' F R2 U' R' U' R U R' F'
x R U R' U' R' F R2 U' R' U' R U R' F'
According to an optimal 2x2 solver, both the scramble that Vincent cuber said as the "next scramble" and the scramble that they solved were both 8 movers.
That can't be possible. If that were the case, the scramble would be a 5 mover, not an 8 mover. That is because all you would have to do is F2 into a 4 mover, which is essentially a 5 mover.
I tried changing the second F into an F' through and it gives a pretty good scramble. There's an obvious 7 mover if you can one look and use cancellation on that scramble. There isn't a 4 move solution though.