Does `str` implement `Borrow<str>`?

Question

I was looking at the Borrow<Borrowed> trait in Rust, and I understand the idea that we want a generic function to allow an argument K or its borrowed form Q such that K: Borrow<Q>. If I look at the String implementation, I see that it implements Borrow<str>, which explains why passing a &String works in the following snippet. What I'm wondering is how can passing a &str work: does that imply that str implements ... Borrow<str>? If not, how can a str satisfy the traits of the given P generic?

use std::borrow::Borrow;
use std::fmt::Display;

fn main() {
    let foo = String::from("I'm a String");
    print(&foo);
    print("I'm a str");
}

fn print<P>(arg: &P) 
where
    P: Borrow<str> + ?Sized + Display
{
    println!("{}", arg)
}

Answer 1

Yes, you can easily look up in the doc

impl<T> Borrow<T> for T where T: ?Sized, 

This means for any type T, Borrow<T> is implemented for T. where T: ?Sized means that T may be dynamically sized. If trait "bound" (more of a relaxation here) where T: ?Sized were not added, then T will not include dynamically sized types like str or other slice types.