If your wild shaped form has damage resistance and you revert to your normal form, does your normal form take the halved damage?

Let’s say we had a druid with the Circle of the Moon archetype. They can use their wild shape feature to turn into an elemental as per their Elemental Wild Shape feature (PHB, p. 69):

Elemental Wild Shape
At 10th level, you can expend two uses of Wild Shape at the same time to transform into an air elemental, an earth elemental, a fire elemental, or a water elemental.

Those elemental creatures all have resistance to nonmagical bludgeoning, piercing and slashing damage, and some forms have other damage resistances too. Let’s say your elemental form was on 5 HP and you took 20 nonmagical slashing damage. Your elemental form would halve that damage to 10 damage. This exceeds the total HP the elemental form was on, and so you would revert back to your normal form, as per wild shape (PHB, p. 67):

… if you revert as a result of dropping to 0 hit points, any excess damage carries over to your normal form. For example, if you take 10 damage in animal form and have only 1 hit point left, you revert and take 9 damage.

Let’s assume that your normal form has no damage resistances. Does your normal form really take just 5 slashing damage, even though your normal form isn’t resistant to slashing damage, or would your normal form take 10 damage, since that’s what it would have been if it wasn’t halved by a resistance that no longer applies to your reverted form? Intuitively I think the normal form should take 5 damage, but the below related question’s answers (even through it’s the opposite scenario) would imply that it would take 10 damage.


Related (but the other way around, since that Q&A is talking about the normal form having the resistance, whereas I’m asking about the wild shaped form having the resistance): How does resistance/vulnerability/immunity interact with carryover damage after reducing Polymorphed (or Wild Shaped) form to 0 HP?