Mixed inequalities with an equality anchor: Given H < J, F < H, and I ≤ J = K, determine which conclusions definitely follow. Conclusions: I) H > I; II) I ≥ F.

Introduction / Context:
This item explores what can (and cannot) be inferred when an element is bounded above by a value equal to another element. The key is recognizing that knowing I ≤ J tells us nothing about I's position relative to H or F unless additional links exist.

Given Data / Assumptions:

• H < J
• F < H
• I ≤ J = K (so I ≤ J and J = K)
• Conclusions to test: I) H > I; II) I ≥ F.

Concept / Approach:
Absence of a direct relation between I and H (or F) leaves multiple possibilities. To disprove a “must follow,” it suffices to build a counterexample consistent with all premises where the conclusion fails.

Step-by-Step Solution:
Counterexample for I (H > I): Let J=10, K=10, H=5 (H < J), I=10 (≤ J). Then H=5 is not > I=10. So I is not forced.Counterexample for II (I ≥ F): Using H=5 and F=1 (F < H), pick I=0 (≤ J). Then I ≥ F is false (0 ≥ 1 fails). Thus II is not forced either.

Verification / Alternative check:
Because both conclusions fail in some consistent assignments, neither is logically necessary. The premises permit a range of values for I relative to H and F.

Why Other Options Are Wrong:
a/b/c/e each claims at least one conclusion must hold. Our counterexamples show neither must.

Common Pitfalls:
Assuming “I ≤ J and H < J” implies I ≥ H or I ≤ H; it implies neither. Without a bridge between I and H, conclusions about their order are indeterminate.

Final Answer:
If neither conclusion I nor II follows