4.12.2 spec says:

Proceeding from left to right, each argument expression e is inferentially typed
...
When a function expression is inferentially typed (section 4.9.3) and a type assigned to a
parameter in that expression references type parameters for which inferences are being made, the
corresponding inferred type arguments to become fixed and no further candidate inferences are
made for them.

But it works differently in compiler. For example:

    function f<T, U>(y: T, f: (x: T) => U, x: T): [T, U] { return [y, f(x)]; }
    interface A { a: A; }
    interface B extends A { b; }

    var a: A, b: B;

    var d = f(b, x => x.a, a); // type [A, A]
    var d2 = f(b, x => x.a, null); // type [B, A]

If "no further candidate inferences were made", then types of d and d2 would be the same. But they are different.