Logical syllogism — determine which conclusions follow beyond doubt. Statements: 1) All the goats are tigers. 2) All the tigers are lions. Conclusions: (1) All the goats are lions. (2) All the lions are goats. (3) Some lions are goats. (4) Some tigers are goats.

Introduction / Context: This is a standard three-set syllogism with nested universals. We must derive all conclusions that must be true from the chain goats ⊆ tigers ⊆ lions.

Given Data / Assumptions:

• All goats are tigers.
• All tigers are lions.
• Non-emptiness of goats for “Some …” conclusions involving existence.

Concept / Approach: Transitivity: if A ⊆ B and B ⊆ C, then A ⊆ C. From “All A are C” and existence of A, we may also infer “Some C are A” and, since A ⊆ B, “Some B are A”.

Step-by-Step Solution: (1) All goats are lions — true by transitivity (A ⊆ C). (2) All lions are goats — converse of (1); not supported. (3) Some lions are goats — since goats exist and every goat is a lion, at least some lions are goats. (4) Some tigers are goats — every goat is a tiger; existence of goats guarantees some tigers are goats.

Verification / Alternative check: Venn diagram with three nested circles immediately shows (1), (3), and (4) as necessary; (2) is not forced and is generally false unless all lions happen to be goats.

Why Other Options Are Wrong: Any option including (2) assumes an unjustified converse.

Common Pitfalls: Confusing subset relations with equality and overlooking the existence requirement for “Some …” conclusions.

Final Answer: Only (1), (3) and (4)