Question about the true significance of the partial trace Consider a composite system whose Hilbert space is $\mathcal{H}_{AB}=\mathcal{H}_A\otimes \mathcal{H}_B$, where $\{|0_A\rangle, |1_A\rangle\}$ and $\{|0_B\rangle, |1_B\rangle\}$ are orthonormal bases of $\mathcal{H}_A$ and $\mathcal{H}_B$, respectively. 
Suppose the composite system is in the pure state $|\psi\rangle = \frac{1}{\sqrt{2}}(|0_A1_B\rangle - |1_A0_B\rangle)$ whose density matrix is given by 
\begin{align}
\rho = \frac{1}{2}(|0_A1_B\rangle \langle 0_A 1_B| + |1_A0_B\rangle \langle 1_A0_B| - |0_A1_B\rangle \langle 1_A0_B| - |1_A0_B\rangle \langle 0_A1_B|)
\end{align}
The state of the $A$-subsystem is then given by the partial trace of $\rho$ with regard to $B$, i.e.
\begin{align}
\rho_A = tr_B(\rho) = \frac{1}{2}(|0_A\rangle \langle 0_A| + |1_A\rangle \langle 1_A|).
\end{align}
$\rho_A$ is a complete description of the $A$-subsystem, and using it we can predict the outcomes of all measurements one can perform on $A$.
Now, taking the partial trace of $\rho$ with regard to $B$ seems to be doing two things: (1) it tells us that the 'off-diagonal terms' in $\rho$ (i.e. the last two terms in the above equation for $\rho$) are irrelevant for the description of the $A$-subsystem; (2) it gives us a way to talk about the $A$-subsystem without mentioning the $B$-basis vectors via the above expression for $\rho_A$. 
However, it seems (to me, at any rate) that (1) is what's really important and significant. Once we know which terms in $\rho$ are required to describe the $A$-subsystem completely, it doesn't matter whether we are mentioning some basis vectors of the $B$-subspace. In other words: it seems as if
\begin{align}
\rho_A' = \frac{1}{2}(|0_A1_B\rangle \langle 0_A 1_B| + |1_A0_B\rangle \langle 1_A0_B|)
\end{align}
is just as good a description of the $A$-subsystem as $\rho_A$; the fact that we're using the language of $\mathcal{H}_{AB}$ is unimportant.
Question: Is this correct? If not, where am I going wrong?
PS: On this line of thought, the operator $\rho_A'$ would also be a complete description of the $B$-subsystem. But this is just a result of the symmetries of the simple example I chose: both partial traces of $\rho$ over $A$ and $B$ enforce the deletion of exactly the same off-diagonal terms. But this is not generally the case.
 A: No, your line of thought is not correct. In fact, it's rather the opposite: we primarily care about the partial trace because of reason (2), i.e., it is the unique way to assign a state to the subsystem A if you do not have any access to measurements in B. The attribute (1) is specific to the particular example you chose and it is essentially a by-product.
If you take two quantum systems with joint state space $\mathcal H_{AB}=\mathcal H_A \otimes \mathcal H_B$, then any pure or mixed state of the joint system can be described by a joint density matrix $\rho_{AB}$, and that joint density matrix can be used to obtain the expectation values of any observable $\hat Q$ via the full trace
$$
\langle Q\rangle = \mathrm{Tr}\mathopen{}\left(\hat Q \rho_{AB}\right)\mathclose{}
.
$$
If you just have access to measurements on subsystem A, though, then you are restricted to measuring observables of the form $\hat Q = \hat R \otimes \mathbb I$, and for those the expectation value simplifies substantially: taking bases $\{|n_A\rangle\}$ and $\{|n_B\rangle\}$ for $\mathcal H_A$ and $\mathcal H_B$, respectively, it reads
\begin{align}
\langle Q\rangle 
& = \mathrm{Tr}\mathopen{}\left(\hat Q \rho_{AB}\right)\mathclose{}
\\ & = \mathrm{Tr}\mathopen{}\left(\hat R\otimes \mathbb I\ \rho_{AB}\right)\mathclose{}
\\ & = \sum_{n_A,n_B}
\langle n_A| \langle n_B|
\left(\hat R\otimes \mathbb I\ \rho_{AB}\right)
|n_A\rangle|n_B\rangle
\\ & = 
\sum_{n_A}
\langle n_A| 
\left(\hat R\ \sum_{n_B}\langle n_B|\mathbb I\rho_{AB}|n_B\rangle\right)
|n_A\rangle
\\ & =
\mathrm{Tr}_A\mathopen{}\left(\hat R \ 
\mathrm{Tr}_B\mathopen{}\left(\rho_{AB}\right)\mathclose{}\right)\mathclose{}
.
\end{align}
In other words, the partial trace $\rho_A=\mathrm{Tr}_B\mathopen{}\left(\rho_{AB}\right)\mathclose{}$ is the density matrix that accounts for all of the experimental observations done on subsystem A that do not involve subsystem B.
Anything that follows from this definition, with the core justification as above, is simply that: a consequence of the definition.
