[dcl.type.auto.deduct]/2 says:
"""
A type T containing a placeholder type, and a corresponding initializer E, are determined as follows:
­— for a variable declared with a type that contains a placeholder type, T is the declared type of the variable and E is the initializer. ...
"""

[dcl.type.auto.deduct]/4 says:
"""
If the placeholder-type-specifier is of the form type-constraint_opt auto, the deduced type T replacing T is determined using the rules for template argument deduction.
Obtain P from T by replacing the occurrences of type-constraint_opt auto either with a new invented type template parameter U, or ... [doesn't matter].
Deduce a value (!!!) for U using the rules of template argument deduction from a function call, where P is a function template parameter type and the corresponding argument is E.
...
[Example:

const auto &i = expr;

The type of i is the deduced type of the parameter u in the call f(expr) of the following invented function template:

template <class U> void f(const U& u);

— end example]
"""

There are several issues here (both in normative wording and the Example): the major one is that for a variable declared as const auto &i = expr;, the initializer is = expr, not expr. And one can't call a function with this initializer as an argument, like f(= expr). Probably, the paragraph should say that the initializer E is the default argument of an invented function template (and the type is deduces as if the function is called without argument)? Nah, the grammar for parameter-declaration doesn't allow initializer.

Also, is it ok that the Standard calls the deduced type "a value of a type parameter" and uses the instructing tone like Obtain P from T ... Deduce a value ....