Merge #6220

6220: implement binary operator overloading type inference r=flodiebold a=ruabmbua

Extend type inference of *binary operator expression*, by adding support for operator overloads.

Before this merge request, the type inference of binary expressions could only resolve operations done on built-in primitive types. This merge requests adds a code path, which is executed in case the built-in inference could not get any results. It resolves the proper operator overload trait in *core::ops* via lang items, and then resolves the associated *Output* type.

```rust
struct V2([f32; 2]);

#[lang = "add"]
pub trait Add<Rhs = Self> {
    /// The resulting type after applying the `+` operator.
    type Output;

    /// Performs the `+` operation.
    #[must_use]
    fn add(self, rhs: Rhs) -> Self::Output;
}

impl Add<V2> for V2 {
    type Output = V2;

    fn add(self, rhs: V2) -> V2 {
        let x = self.0[0] + rhs.0[0];
        let y = self.0[1] + rhs.0[1];
        V2([x, y])
    }
}

fn test() {
    let va = V2([0.0, 1.0]);
    let vb = V2([0.0, 1.0]);

    let r = va + vb; // This infers to V2 now
}
```

There is a problem with operator overloads, which do not explicitly set the *Rhs* type parameter in the respective impl block. 

**Example:**

```rust
impl Add for V2 {
    type Output = V2;

    fn add(self, rhs: V2) -> V2 {
        let x = self.0[0] + rhs.0[0];
        let y = self.0[1] + rhs.0[1];
        V2([x, y])
    }
}
```

In this case, the trait solver does not realize, that the *Rhs* type parameter is actually self in the context of the impl block. This stops type inference in its tracks, and it can not resolve the associated *Output* type.

I guess we can still merge this back, because it increases the amount of resolved types, and does not regress anything (in the tests).

Somewhat blocked by https://github.com/rust-analyzer/rust-analyzer/issues/5685
Resolves  https://github.com/rust-analyzer/rust-analyzer/issues/5544

Co-authored-by: Roland Ruckerbauer <roland.rucky@gmail.com>

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