I don't understand what you're saying. Are you saying that "lazy borrowing" is how the compiler works, or that it's just how you mentally model Rust for your own convenience? There is no "lazy borrowing" in the Rust compiler, as far as I know.

You can create a mutable reference `mut1`, mutate through it, create a new mutable reference `mut2`, mutate through it, then mutate through `mut1` again. Here is an example to illustrate:

  fn main() {
    let mut s = String::from("1"); // create value
  
    let mut1: &mut _ = &mut s; // create a ref mut
    *mut1 = String::from("2"); // write through mut1
    println!("using mut1: {mut1}"); // print mut1
  
    let mut2: &mut _ = &mut *mut1; // create a second ref mut
    *mut2 = String::from("3"); // write through mut2
    println!("using mut2: {mut2}"); // print mut2
    
    *mut1 = String::from("4"); // write through mut1 again
    println!("using mut1: {mut1}"); // print mut1
  }

Lifetime analysis can infer shorter lifetimes for unused refs. But creating refs in the wrong order matters even if you never use them, and not all orders are valid Rust programs. Here's a very small example that doesn't compile:

  fn main() {
    let mut s = String::from("1");

    let mut1: &mut _ = &mut s;
    let mut2: &mut _ = &mut s; // never used!
  
    println!("Using mut1: {mut1}");
  }