[intro.execution] p3 says

The immediate subexpressions of an expression E are

• the constituent expressions of E's operands ([expr.prop]),
• any function call that E implicitly invokes,
• if E is a lambda-expression ([expr.prim.lambda]), the initialization of the entities captured by copy and the constituent expressions of the initializer of the init-captures,
• if E is a function call or implicitly invokes a function, the constituent expressions of each default argument ([dcl.fct.default]) used in the call, or
• if E creates an aggregate object ([dcl.init.aggr]), the constituent expressions of each default member initializer ([class.mem]) used in the initialization.

A subexpression of an expression E is an immediate subexpression of E or a subexpression of an immediate subexpression of E.

Consider this example:

int a = 0, b = 1;
a+b;  // #1

[basic.lval] p6 says

Whenever a glvalue appears as an operand of an operator that expects a prvalue for that operand, the lvalue-to-rvalue, array-to-pointer, or function-to-pointer standard conversions are applied to convert the expression to a prvalue.

So, any of the operands at #1 is the prvalue rather than a or b. According to the first bullet, such two prvalues are the operands. The constituent expression of the prvalue is itself. That means a and b are not subexpressions of the expression at #1. However, if the conversion is a user-defined conversion, the associated function call is a subexpression, as per the second bullet.

Moreover, this can introduce a doubt, when we say

the full-expression of its initialization is a constant expression when interpreted as a constant-expression

What can be subexpressions of the constant-expression when built-in conversion applies?

const int v = 0;
constexpr int i = v;

The initialization should be the prvalue that results from applying lvalue-to-rvalue conversion to v, In analogy to above, v is not a subexpression in that constant-expression.