OFB output jth-block expression clarification

Question

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?

Answer 1

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:

$$C_{j-1} \oplus P_{j-1} = O_{j-1}$$