Statements: 1) All goats are tigers. 2) All tigers are lions. Conclusions: I. All tigers are goats. II. All lions are tigers. III. No goat is a lion. IV. No lion is a goat. Choose the option that must follow.

Introduction / Context:A pair of chained universal affirmatives places one class inside another and then inside a third. The conclusions tempt you with converses and absolute negatives. We must evaluate each against the actual chain.

Given Data / Assumptions:

• Goats ⊆ Tigers.
• Tigers ⊆ Lions.
• Therefore, Goats ⊆ Lions (by transitivity).

Concept / Approach:From A ⊆ B and B ⊆ C, we may derive A ⊆ C, but we cannot reverse them to B ⊆ A or C ⊆ B. Also, “No A are C” is directly contradicted when we have A ⊆ C unless A is empty (which is not asserted here).

Step-by-Step Solution:1) Conclusion I (“All tigers are goats”) is the converse of “All goats are tigers”; it is not entailed.2) Conclusion II (“All lions are tigers”) is the converse of “All tigers are lions”; not entailed.3) Conclusion III (“No goat is a lion”) contradicts the chain Goats ⊆ Lions; unless Goats were empty (not stated), this is false.4) Conclusion IV (“No lion is a goat”) is also false because at least those goats are lions; hence some lions are goats.

Verification / Alternative check:Model: Let Goats = {g1}, Tigers = {g1, t2}, Lions = {g1, t2, l3}. Both premises hold. I fails (t2 is a tiger but not a goat), II fails (l3 is a lion but not a tiger), III fails (g1 is a goat and a lion), IV fails (g1 is a lion and a goat). Hence none of the conclusions follows.

Why Other Options Are Wrong:Options claiming any of I–IV follow contradict either the chain or commit converse errors.

Common Pitfalls:Reversing subset statements and assuming disjointness where inclusion is given.

Final Answer:None of the Conclusions follows.