Statements: S ≥ Y ≥ E = F ≥ R > P = L Conclusions: F < S E > L

Concept

In statement-and-conclusion reasoning, chaining two ‘≥’ links, or a ‘≥’ with an ‘=’, keeps the combined relation as ‘≥’ — equality along the whole path is still possible. But chaining a ‘≥’ with even one strict ‘>’ anywhere along the path makes the combined relation strict. A conclusion between two terms is definitely true only when the chain connecting them actually forces that relation — if an equality is possible anywhere on the path, a strict conclusion between those terms is not guaranteed.

Applying it to this chain

1. Write the full given chain: S ≥ Y ≥ E = F ≥ R > P = L.

2. Conclusion I compares S and F. The path from S to F is S ≥ Y ≥ E = F — only ‘≥’ and ‘=’ links, no strict link anywhere. So the forced relation is S ≥ F, not S > F; S = Y = E = F is consistent with every stated relation.

3. Conclusion II compares E and L. The path from E to L is E = F ≥ R > P = L — this path contains the strict link R > P. Chaining ‘=’, then ‘≥’, then ‘>’ gives E ≥ R > L, an overall strict relation.

4. So Conclusion I (F < S) is not guaranteed (equality is possible), while Conclusion II (E > L) is guaranteed in every valid arrangement.

5. Only Conclusion II holds — this matches the option stating exactly that.

Cross-check

Test a boundary assignment that satisfies every stated relation: S = Y = E = F = R = 4, P = L = 3. Check the statement: 4 ≥ 4 ≥ 4 = 4 ≥ 4 > 3 = 3 — all relations hold. Here F < S becomes 4 < 4, which is false, confirming Conclusion I can fail. In this same assignment E > L is 4 > 3, which holds, confirming Conclusion II holds even at this boundary — and it holds in every other valid arrangement too, since it follows directly from the chain.