I've posted similar minimal examples before. The first one triggers a move (and then fails to compile), while the second one triggers a reborrow. Sure, you can explicitly write out every instance of reborrowing with `&mut *`. That would require you to understand every instance that triggers it, and would also be unbelievably noisy, since automatic dereferencing and automatic reborrowing are actually incredibly common.

  > pub fn main() {
  >     let mut x = String::from("moo");
  >     let y = &mut x;
  >     let z = y;
  >     println!("{:?}", y)
  > }
  > 
  > pub fn main() {
  >     let mut x = String::from("moo");
  >     let y = &mut x;
  >     let z: &mut _ = y;
  >     println!("{:?}", y)
  > }