# Reduction from Real-Or-Random to Left-Or-Right

I am reading the paper A Concrete Security Treatment of Symmetric Encryption and am confused by the reduction from ROR to LOR on page 11. Specifically, when it says:

When $$\mathcal{O}_2(\cdot)=\mathcal{E}_K(\mathcal{RR}(\cdot,0))$$, we have that $$\mathcal{O}_1(\mathcal{LR}(\cdot,\cdot,0))$$ and $$\mathcal{O}_1(\mathcal{LR}(\cdot,\cdot,1))$$ return identically distributed answers.

So, $$\Pr[\mathbf{Exp}_\mathcal{SE,A_2}^{ror-atk-0}(k)=1]=1/2$$

I'm not sure how the first statement implies the second statement.

A key aspect is that whenever we are in experiment $$\text{ror-atk-}0$$, $$\mathcal{A}_1$$ receives a ciphertext $$c$$ corresponding to a random plaintext. In particular, this random plaintext is (by definition) independent of the bit $$b$$ selected by the $$\text{lor}$$ adversary $$\mathcal{A}_2$$. In turn, $$\mathcal{A}_1$$'s guess $$d$$ is also independent of $$b$$ since it only depends on the ciphertext for random plaintexts and the random choices made by $$\mathcal{A}_1$$. Hence, the probability expression is equivalent to $$\Pr[\mathbf{Exp}_\mathcal{SE,A_2}^{ror-atk-0}(k)=1]= \Pr[d = b] = 1/2.$$