Test a Syllogism with Sets and Counterexamples

Understand the concept

A syllogism asks whether a conclusion necessarily follows from stated premises. Work only with those premises. A conclusion is not valid merely because it is often true in everyday life or seems compatible with the statements.

Represent “all A are B” by placing the A set entirely inside the B set. This does not require the two sets to be equal. “No A are B” means their overlap is empty. “Some A are B” means at least one member lies in the overlap. In modern set logic, an all-statement by itself does not guarantee that any A members exist; an explicit some-statement does make an existence claim.

Containment can be chained. If all A are B and all B are C, every A member is also a C member, so all A are C follows. The reverse conclusion, all C are A, does not follow. There may be C members outside B or B members outside A.

Shared membership in a broad set does not prove overlap between two smaller sets. If all poets are writers and all historians are writers, poets and historians might overlap, but the premises do not force them to do so. They could occupy separate parts of the writer set.

To disprove a proposed conclusion, construct one counterexample in which every premise is true and the conclusion is false. A counterexample is stronger than saying that a conclusion feels uncertain: it demonstrates a specific permitted arrangement. To establish that a conclusion follows, show why the premises rule out every arrangement that would make it false.

Keep existence visible. If some A are B and all B are C, the particular member in the A–B overlap must also be in C, so some A are C follows. By contrast, all A are B and all B are C alone do not establish that some A are C, because A could be empty under the stated convention.

Worked example

Premises: All squares are rectangles. Some rectangles are blue. Proposed conclusion: Some squares are blue.

Draw the square set inside the rectangle set. The “some rectangles are blue” statement requires at least one blue member somewhere in the rectangle set, but it does not say that the member is a square.

A counterexample places the required blue member among rectangles that are not squares and leaves all squares non-blue. Both premises remain true, while the conclusion is false. Therefore the conclusion does not follow.

Now change the second premise to “Some squares are blue.” Together with “All squares are rectangles,” it does follow that some rectangles are blue: the explicitly existing blue square is also a rectangle.

Practice and answer

1. All A are B. No B are C. Does “No A are C” follow?
Answer: yes. Since A lies within B and B has no overlap with C, A cannot overlap C.

2. Some A are B. All B are C. Does “Some A are C” follow?
Answer: yes. The member guaranteed in the A–B overlap must also belong to C.

3. All A are B. All C are B. Does “Some A are C” follow?
Answer: no. A and C can occupy separate parts of B, and the all-statements alone do not require either set to contain a member.

4. All A are B. Does “All B are A” follow?
Answer: no. B may contain additional members outside A.

What you will learn

• Translate all, none and some statements into set relationships.
• Distinguish a possible conclusion from a necessary one.
• Use a counterexample and handle existence explicitly.