Does the standard mandate an lvalue-to-rvalue conversion of the pointer variable when applying indirection?

I think you're approaching this from a rather oblique angle, so to speak. According to §5.3.1/1:

The unary * operator performs indirection: the expression to which it is applied shall be a pointer to an object type, or a pointer to a function type and the result is an lvalue referring to the object or function to which the expression points. If the type of the expression is “pointer to T,” the type of the result is “T.”

Although this doesn't talk about the lvalue-to-rvalue conversion, it requires that the expression be a pointer to an object or function. An uninitialized pointer won't (except, perhaps by accident) be any such thing so the attempt at dereferencing gives undefined behavior.

I have converted the update section in my question to an answer since at this point it seems to be the answer, albeit an unsatisfactory one that my question is unanswerable:

dyp pointed me to two relevant threads that cover very similar ground:

• What is the value category of the operands of C++ operators when unspecified?
• Does initialization entail lvalue-to-rvalue conversion? Is int x = x; UB?

The consensus seems to be that the standard is ill-specified and therefore can not provide the answer I am looking for, Joseph Mansfield posted a defect report on this lack of specification, and it looks like it is still open and it is not clear when it may be clarified.

There are a few common sense arguments to be made as to the intent of the standard. One can argue Logicially, an operand is a prvalue if the operation requires using the value of that operand. Another argument is that if we look back to the C99 draft standard says an lvalue to rvalue conversion is done by default and the exceptions are noted. The relevant section from the draft C99 standard is 6.3.2.1 Lvalues, arrays, and function designators paragraph 2 which says:

Except when it is the operand of the sizeof operator, the unary & operator, the ++ operator, the -- operator, or the left operand of the . operator or an assignment operator, an lvalue that does not have array type is converted to the value stored in the designated object (and is no longer an lvalue). […]

which basically says with some exceptions an operand is converted to the value stored and since indirection is not an exception if this is clarified to also be the case in C++ as well then it would indeed make the answer to my question yes.

As I attempted to clarify the proof of undefined behavior was less important than clarifying whether a lvalue-to-rvalue conversion is mandated. If we want to prove undefined behavior we have alternate approaches. Jerry’s approach is a common sense one and in that indirection requires that the expression be a pointer to an object or function and an indeterminate value will only by accident point to a valid object. In general the draft C++ standard does not give an explicit statement to say using an indeterminate value is undefined, unlike the C99 draft standard In C++11 and back the standard does not give an explicit statement to say using an indeterminate value is undefined. The exception being iterators and by extension pointers we do have the concept of singular value and we are told in section 24.2.1 that:

[…][ Example: After the declaration of an uninitialized pointer x (as with int* x;), x must always be assumed to have a singular value of a pointer. —end example ] […] Dereferenceable values are always non-singular.

and:

An invalid iterator is an iterator that may be singular.²⁶⁸

and footnote 268 says:

This definition applies to pointers, since pointers are iterators. The effect of dereferencing an iterator that has been invalidated is undefined.

In C++1y the language changes and we do have an explicit statement making the use of an intermediate value undefined with some narrow exceptions.