If ABR is a pure state, why S(A,B) = S(R), where S is the Von Neumann entropy? Let A,B be two distinct quantum systems that have a joint state $\rho ^{AB}$. If R is a system that purifies A,B why, by virtue that ABR is a pure state, S(A,R) = S(B) and S(A,B) = S(R)?
I came across the above at a step in the proof of the triangle inequality $ S(A,B) \geq \left | S(A) - S(B) \right |$ in Nielsen & Chuang's text, section 11.3.4 (p516 in 1st edition). They do not give any further explanation.
 A: For any pure state $S$ that you separate into two (possibly entangled) pieces $A$ and $B$, you can use the Schmidt decomposition to rewrite the state into the form:
$$S = \sum_k s_k |a_k\rangle |b_k\rangle$$
where the $s_k$ scalars are between 0 and 1 and have squares that sum up to 1, the $a_k$ vectors are from the $A$ part while being mutually perpendicular, and the $b_k$ vectors are from the $B$ part while being mutually perpendicular.
When you trace out $B$, the mutually perpendicular $|b_k\rangle$ vectors effectively split the state into the mixed state
$$S_A = \sum_k s_k^2 |a_k\rangle \langle a_k|$$
Similarly
$$S_B = \sum_k s_k^2 |b_k\rangle \langle b_k|$$
In other words, when focusing on just $A$ there's a $s_0^2$ probability of being in state $|a_0\rangle$, a $s_1^2$ probability of being in state $|a_1\rangle$, and so forth.
Since the $s_k$ values define the probabilities, and the Von Neumann entropy is a function of the probabilities, and both sums use the same $s_k$, the entropy of $S_A$ is equal to the entropy of $S_B$.
It doesn't matter whether you make overall pure state via a purification step, or make various different decisions on where to place the division between the two subsystems. This works for any pure state and any division.
