Given that All A are B and Some B are C, can we conclude Some A are C?

Given that All A are B and Some B are C, can we conclude Some A are C?

• A. Yes
• B. Only if B and C overlap
• C. Impossible
• D. Not necessarily; cannot deduce

Answer: (D)

This question tests how a subset relationship and an existence claim interact. From All A are B we know A sits inside B, so every A is a B. From Some B are C we know there exists at least one thing that is both B and C. But that thing doesn’t have to be in A, because A could be just a part of B, and C could intersect B outside of A. A concrete way to see it: take A = {1}, B = {1, 2}, C = {2}. All A are B holds (since 1 is in B), and some B are C holds (2 is in both B and C), yet some A are C is false because A contains only 1, which is not in C. On the other hand, if we choose A = {1, 2}, B = {1, 2, 3}, C = {2}, then all A are B holds, some B are C holds, and some A are C also holds (2 is in A and C). Because the truth of some A are C depends on how A, B, and C actually overlap, the conclusion cannot be guaranteed. Not necessarily; cannot deduce.

This question tests how a subset relationship and an existence claim interact. From All A are B we know A sits inside B, so every A is a B. From Some B are C we know there exists at least one thing that is both B and C. But that thing doesn’t have to be in A, because A could be just a part of B, and C could intersect B outside of A.

A concrete way to see it: take A = {1}, B = {1, 2}, C = {2}. All A are B holds (since 1 is in B), and some B are C holds (2 is in both B and C), yet some A are C is false because A contains only 1, which is not in C.

On the other hand, if we choose A = {1, 2}, B = {1, 2, 3}, C = {2}, then all A are B holds, some B are C holds, and some A are C also holds (2 is in A and C).

Because the truth of some A are C depends on how A, B, and C actually overlap, the conclusion cannot be guaranteed. Not necessarily; cannot deduce.