So there is a phrase from the C++ Standard working draft (n4928):

[dcl.init.general] ¶ 16.6.1

If the initializer expression is a prvalue and the cv-unqualified version of the source type is the same as the destination type, the initializer expression is used to initialize the destination object [Example 2: T x = T(T(T())); value-initializes x. — end example]

For me, the words "is used to initialize" sounds like an infinite recursion of "the process of initialization" mentioned in [dcl.init.general] ¶ 1. Yes, I re-read carefully [basic.lval] ¶ 5 and [expr.type.conv] and understand now why the reader of the Standard is supposed to think that neither copy- nor move-constructors of T are called. But again the wording in [dcl.init.general] ¶ 16.6.1 could be improved like this:

[dcl.init.general] ¶ 16.6.1 (corrected)

If the initializer expression is a prvalue and the cv-unqualified version of the source type is the same as the destination type, the result object of the initializer expression is the object being initialized. [Example 2: T x = T(T(T())); x is value-initialized according to [expr.type.conv] and [basic.lval]. — end example]

[dcl.init.general] ¶ 16.9 (corrected)

Otherwise, the result object of the (possibly converted) initializer expression is the object being initialized.

[basic.lval] ¶ 5 (corrected)

...The result object of a prvalue is the object initialized by the prvalue and the initial value of which is that of the prvalue; (italicized words - sic!)...

Another (minimal change) variant is to add a few words to the example in:

[dcl.init.general] ¶ 16.6.1 (corrected)

...the initializer expression is used to initialize the destination object. [Example 2: T x = T(T(T())); x is value-initialized according to [expr.type.conv] and [basic.lval]. — end example]

This part of the Standard is important, because it makes mandatory the so called unnamed return value optimization (URVO) for compilers since C++17...

Discussion is also on Stack Overflow here.