Discussion
Statements: S ≤ T ≤ K ≤ E; N ≥ E ≤ P
Conclusions:
I. S ≤ N
II. S ≤ P
Answer: E
Correct answer: Both conclusions I and II are true.
The statements are \( S \le T \le K \le E \) and \( N \ge E \le P \). E is the junction: the first chain climbs to it from S, and both N and P stand at or above it.
I. \( S \le N \): the run \( S \le T \le K \le E \le N \) points one way at every link, so \( S \le N \). True.
II. \( S \le P \): the same run with P in place of N - \( S \le T \le K \le E \le P \). True.
Both follow. What makes this one work where so many others fail is that the two branches leaving E both point upwards, in the same direction as the chain arriving at it, so E is not a pivot and the relations carry straight through. The comparison this arrangement would still refuse is N against P - those two are the pair on separate branches above a common element.
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