Strictly speaking, the initial part of your assumption is not wrong, but the later conclusion is.
array[x] and *(array+x) are indeed equivalent for any identifiers 'array' and 'x' (assuming one of those evaluates as a pointer value, and the other as an integer value; otherwise the code is incorrect). In fact, in this context an actual array is not subject to either unary-& or sizeof operators, so it evaluates to a pointer value, fulfilling the precondition.
This is why "array subscription" also directly works with pointers (i.e. "ptr[x]"), and from the equivalence above follows one of the common useless facts that you can swap the identifiers (i.e. "x[array]").
(This comment is probably confusing enough without saying that "(&array)[x]" is valid code too, but isn't the same thing as those before.)
array[x] and *(array+x) are indeed equivalent for any identifiers 'array' and 'x' (assuming one of those evaluates as a pointer value, and the other as an integer value; otherwise the code is incorrect). In fact, in this context an actual array is not subject to either unary-& or sizeof operators, so it evaluates to a pointer value, fulfilling the precondition.
This is why "array subscription" also directly works with pointers (i.e. "ptr[x]"), and from the equivalence above follows one of the common useless facts that you can swap the identifiers (i.e. "x[array]").
(This comment is probably confusing enough without saying that "(&array)[x]" is valid code too, but isn't the same thing as those before.)