Mixed non-strict/strict chain: Given H ≥ I = J > K ≤ L, determine which conclusions definitely follow. Conclusions: I) K < H; II) L ≥ I.

Introduction / Context:
This problem asks what must follow from a chain mixing ≥, =, >, and ≤. We compare K against H and L against I carefully, watching where strictness is preserved or lost.

Given Data / Assumptions:

• H ≥ I = J
• J > K
• K ≤ L
• Conclusions: I) K < H; II) L ≥ I.

Concept / Approach:
From I = J and J > K, we know I > K. Combining H ≥ I with I > K yields H > K ⇒ K < H (I true). For II, K ≤ L and I > K do not force L ≥ I; L might sit between K and I.

Step-by-Step Solution:
I) Since I = J and J > K ⇒ I > K. With H ≥ I, we have H > K ⇒ K < H (true).II) From K ≤ L and I > K, L could be ≤ I or < I; nothing mandates L ≥ I. Thus II is not forced.

Verification / Alternative check:
Counterexample for II: Let K=2, L=3, I=5, H=6 (and J=5). Then L ≥ I is false (3 ≥ 5 fails), while all premises hold.

Why Other Options Are Wrong:
b/d/c each claim a truth value for II that is not compelled; only I is guaranteed.

Common Pitfalls:
Assuming K ≤ L and I > K implies L ≥ I—this is not valid without a link between L and I beyond the shared comparison with K.

Final Answer:
If only Conclusions I is true