When you first see both A - B and A \cap B' in the same chapter, the reflex is to assume they must mean something different — why else would the textbook introduce two notations? Then the worked examples give the same answer for both and the doubt sharpens. Is there a subtle distinction, or are these genuinely two names for the same object?

The answer: they are the same set, always. A - B is defined as "elements of A not in B," and A \cap B' unpacks to "elements of A and not in B." Same condition, same set. The two notations survive because each one is natural in a different context.

The definitions, side by side

A - B = \{ x \mid x \in A \text{ and } x \notin B \}
A \cap B' = \{ x \mid x \in A \text{ and } x \in B' \}

Now observe that "x \in B'" means exactly "x \notin B" (that is the definition of complement). Substitute:

A \cap B' = \{ x \mid x \in A \text{ and } x \notin B \} = A - B

The two set-builder conditions are word-for-word identical. Whatever x is, the statement "x \in A - B" is true exactly when "x \in A \cap B'" is true. Same truth value everywhere means same set.

A one-line proof

Claim. A - B = A \cap B' for all sets A, B inside a universal set U.

Proof. For any x,

x \in A - B \iff (x \in A \text{ and } x \notin B) \iff (x \in A \text{ and } x \in B') \iff x \in A \cap B'.

Therefore A - B = A \cap B'. \blacksquare

Every step is a definition — no clever manipulation needed.

Worked example

Let U = \{1, 2, 3, 4, 5, 6, 7, 8\}, A = \{1, 2, 3, 4, 5\}, B = \{4, 5, 6, 7\}.

Compute A - B directly:

A - B = \{1, 2, 3\}

Why: go through A and keep only the elements that are not in B. The elements 4 and 5 appear in B, so they get dropped. What remains is \{1, 2, 3\}.

Now compute A \cap B':

B' = U - B = \{1, 2, 3, 8\}
A \cap B' = \{1, 2, 3, 4, 5\} \cap \{1, 2, 3, 8\} = \{1, 2, 3\}

Why: first find everything in the universe not in B — that is B'. Then keep only the elements of A that are also in B'. The intersection of A and B' is \{1, 2, 3\}.

Same answer. Not a coincidence — guaranteed by the identity.

On the Venn diagram they are the same region

Two Venn diagrams showing A minus B and A intersect B complement as identical regionsTwo Venn diagrams side by side each containing two overlapping circles A and B inside a rectangle. On the left the crescent part of A that is outside B is shaded indicating A minus B. On the right the same crescent region is shaded indicating A intersect B complement. An equals sign sits between the two diagrams. A − B A B = A ∩ B′ A B
The left crescent is the same region in both diagrams. On the left, it is described as "inside $A$, not inside $B$" — $A - B$. On the right, it is described as "inside $A$ and inside $B'$" — $A \cap B'$. Different descriptions, same region.

Why two notations, then?

Because each one matches a different way of thinking.

A - B fits when you think of removal. "Start with A, then take away the elements that happen to be in B." The dash evokes subtraction in arithmetic — and indeed, for finite disjoint sets, |A - B| = |A| - |A \cap B| looks a lot like numerical subtraction.

A \cap B' fits when you think of conjunction. "Elements that are in A and are in B'." This form makes it obvious that A - B is a special case of intersection, which is important when you want to apply identities of Boolean algebra. De Morgan, distributive laws, and complement rules are all stated in terms of \cap, \cup, and ' — not in terms of -. To use those identities on a set difference, you first rewrite it as A \cap B'.

Example. Suppose you want to prove (A - B) \cup (A - C) = A - (B \cap C). Rewrite both sides using \cap and ':

Both sides equal A \cap (B' \cup C'). The identity holds.

That proof is painless in the \cap-' form and awkward in the - form. But the statement reads more naturally in the - form. The two notations are complementary tools, and mature set theory uses whichever is clearer in the moment.

When the universe matters

The complement B' depends on the universal set U. Change U and B' changes. But the set difference A - B does not depend on U — it is defined purely in terms of A and B.

So if you are working in a context with no specified universe, A - B is always well-defined, while A \cap B' requires you to commit to a universe first. This is another reason both notations exist: A - B is the "universe-free" version, and A \cap B' is the "universe-aware" version. For the subset of B that actually sits inside A, the two give the same answer regardless.

Three statements that always hold

These are identities that connect the two forms. They are worth recognising at a glance.

Every one is provable by a one-line chase of membership. When a problem gives A - B and an algebraic identity in the \cap-' language, the identity A - B = A \cap B' is the bridge.

So what is the student's doubt actually pointing at?

When you first meet A - B and A \cap B' separately, you are pattern-matching on notation. Two different symbol-groups feel like they must describe two different things. The realisation that they don't is part of a broader lesson in set theory: the same set can have many names. \{x \in \mathbb{R} \mid x^2 < 1\}, (-1, 1), and \{x \in \mathbb{R} \mid -1 < x < 1\} all name the same interval; A - B and A \cap B' all name the same region. Sets are identified by what they contain, not by the formula used to describe them.

Once you internalise this, the notation stops being a barrier. You pick whichever form makes the current argument shortest.

Related: Set Operations · Three-Set Venn Diagram · Symmetric Difference · Sets — Introduction