Ordered relations: A ≥ B = C and B < D ≤ E. Which conclusion(s) definitely follow? I) D > A; II) E > C.

Introduction / Context:
We combine non-strict and strict relations from two chains meeting at B = C to deduce consequences for D and E relative to A and C.

Given Data / Assumptions:

• A ≥ B = C
• B < D ≤ E

Concept / Approach:
Use transitivity from B = C into the second chain and compare against A via A ≥ C.

Step-by-Step Solution:
Since B = C and B < D, we get C < D. With D ≤ E, we have E ≥ D > C ⇒ E > C (II true). For I, A ≥ C while D > C. A could be greater than, equal to, or less than D (no direct link). So D > A is not compelled (I not forced).

Verification / Alternative check:
Counterexample for I: let C=B=5, A=9, D=6, E≥6. Then D > A is false though all premises hold.

Why Other Options Are Wrong:
Only II is guaranteed; the others assert more than the premises entail.

Common Pitfalls:
Assuming A is tied to D because both relate to C; without A vs D linkage, I is undetermined.

Final Answer:
If only conclusion II follows