Note: for readers of the 2016 LGB, this game appears on page 359.
Thanks for the question! While you are correct that at least one of J and S must be in the forest, that rule also allows for both J and S to be in the forest. So, let's take a look at what can occur in this question.
The fourth rule in this game states that, "If jays are not in the forest, then shrikes are." The placement of the "not" here is critical, because it's not where it usually is. In this case, it's on the sufficient condition, not the necessary condition. Thus, the diagram is:
As detailed on page 268 and 269 in your edition of the book (side note: for readers of the 2016 edition this discussion appears on pages 294-295), this presentation—where the sufficient condition is negative but not the necessary—is dangerous. Many students take this rule to mean that J and S can't be in the forest together, but the meaning is actually the polar opposite. The meaning of the rule as given comes down to: when one of J or S is absent, the other must be there. Thus, at least one of J or S is always selected, possibly both. Thus, under the rule there are three possible outcomes:
- 1. S not in forest, J in forest
2. J not in forest, S in forest
3. Both J and S in forest
That last possibility occurs because if either J or S is selected, that does not violate any condition in the J
So, in question #7, when we get down to M and H are in, and that G and W are out, only J and S are left, and you could have either one or both. Hence, answer choice (E) is correct, and the "at most" includes the possibility of four birds.
Here's the good news: now that you've seen this "negative sufficient condition" rule presentation, you won't fall for it again. And that's one of the great values of preparing—you get a chance to see what they can do and how you should best handle it. This rule type has been appearing with greater frequency recently, and so it's important to know how tricky it can be.
Please let me know if that helps. Thanks!