OFB output jth-block expression clarification Stalling says:

OFB encryption can be expressed as
Cj = Pj⊕E(K, Oj-1)
where
Oj-1 = E(K, Oj-2)
Some thought should convince you that we can rewrite the encryption
expression as:
Cj = Pj⊕E(K, [Cj-1 ⊕Pj-1])

Is he implicitly saying that

[Cj-1 ⊕Pj-1] = Oj-1

? If yes, why?
Is he implicitly saying that
[Cj-1 ⊕Pj-1] = Oj-1

Yes, this is true.

? If yes, why?

Well, we know that:
$$C_{j-1} = P_{j-1} \oplus E(K, O_{j-2})$$
(this is the first formula, replacing $j$ with $j-1$), and we know that:
$$O_{j-1} = E(K, O_{j-2})$$
Combining the two, we get:
$$C_{j-1} = P_{j-1} \oplus O_{j-1}$$
or, in other words:
$$C_{j-1} \oplus P_{j-1} = O_{j-1}$$