$\frac{P(x_1 \mid y, s = 1) \dots P(x_n \mid y, s = 1) P(y \mid s = 1)}{P(x \mid s = 1)}$ indicates that naive Bayes learners are global learners?

I am currently studying the paper Learning and Evaluating Classifiers under Sample Selection Bias by Bianca Zadrozny. In section 3. Learning under sample selection bias, the author says the following:

We can separate classifier learners into two categories:

• local: the output of the learner depends asymptotically only on $$P(y \mid x)$$
• global: the output of the learner depends asymptotically both on $$P(x)$$ and on $$P(y \mid x)$$.

The term "asymptotically" refers to the behavior of the learner as the number of training examples grows. The names "local" and "global" were chosen because $$P(x)$$ is a global distribution over the entire input space, while $$P(y \mid x)$$ refers to many local distributions, one for each value of $$x$$. Local learners are not affected by sample selection bias because, by definition $$P(y \mid x, s = 1) = P(y \mid x)$$ while global learners are affected because the bias changes $$P(x)$$.

Then, in section 3.1.1. Naive Bayes, the author says the following:

In practical Bayesian learning, we often make the assumption that the features are independent given the label $$y$$, that is, we assume that $$P(x_1, x_2, \dots, x_n \mid y) = P(x_1 \mid y) P(x_2 \mid y) \dots P(x_n \mid y).$$ This is the so-called naive Bayes assumption. With naive Bayes, unfortunately, the estimates of $$P(y \mid x)$$ obtained from the biased sample are incorrect. The posterior probability $$P(y \mid x)$$ is estimated as $$\dfrac{P(x_1 \mid y, s = 1) \dots P(x_n \mid y, s = 1) P(y \mid s = 1)}{P(x \mid s = 1)} ,$$ which is different (even asymptotically) from the estimate of $$P(y \mid x)$$ obtained with naive Bayes without sample selection bias. We cannot simplify this further because there are no independence relationships between each $$x_i$$, $$y$$, and $$s$$. Therefore, naive Bayes learners are global learners.

Since it is said that, for global learners, the output of the learner depends asymptotically both on $$P(x)$$ and on $$P(y \mid x)$$, what is it about $$\dfrac{P(x_1 \mid y, s = 1) \dots P(x_n \mid y, s = 1) P(y \mid s = 1)}{P(x \mid s = 1)}$$ that indicates that naive Bayes learners are global learners?

EDIT: To be clear, if we take the example given for the local learner case (section 3.1. Bayesian classifiers), then it is evident:

Bayesian classifiers compute posterior probabilities $$P(y \mid x)$$ using Bayes' rule: $$P(y \mid x) = \dfrac{P(x \mid y)P(y)}{P(x)}$$ where $$P(x \mid y)$$, $$P(y)$$ and $$P(x)$$ are estimated from the training data. An example $$x$$ is classified by choosing the label $$y$$ with the highest posterior $$P(y \mid x)$$.

We can easily show that bayesian classifiers are not affected by sample selection bias. By using the biased sample as training data, we are effectively estimating $$P(x \mid y, s = 1)$$, $$P(x \mid s = 1)$$ and $$P(y \mid s = 1)$$ instead of estimating $$P(x \mid y)$$, $$P(y)$$ and $$P(x)$$. However, when we substitute these estimates into the equation above and apply Bayes' rule again, we see that we still obtain the desired posterior probability $$P(y \mid x)$$: $$\dfrac{P(x \mid y, s = 1) P(y \mid s = 1)}{P(x \mid s = 1)} = P(y \mid x, s = 1) = P(y \mid x)$$ since we are assuming that $$y$$ and $$s$$ are independent given $$x$$. Note that even though the estimates of $$P(x \mid y, s = 1)$$, $$P(x \mid s = 1)$$ and $$P(y \mid s = 1)$$ are different from the estimates of $$P(x \mid y)$$, $$P(x)$$ and $$P(y)$$, the differences cancel out. Therefore, bayesian learners are local learners.

Note that we get $$P(y \mid x)$$. However, in the global case, it is not clear how we get $$P(x)$$ and $$P(y \mid x)$$ (as is required for global leaners) from $$\dfrac{P(x_1 \mid y, s = 1) \dots P(x_n \mid y, s = 1) P(y \mid s = 1)}{P(x \mid s = 1)}$$.