First time you see De Morgan's first law, (A \cup B)' = A' \cap B', the symbols look more intimidating than the idea deserves. Complement, union, complement, intersection, complement — five operators in six characters. A lot of students memorise it as a rule and hope to recognise when to apply it. The memorisation is unnecessary; the intuition is a single plain-English sentence.

"Not in either" is the same as "not in the first and not in the second."

That sentence is the whole law. Everything that follows is just repackaging it in different notation.

The English translation, slowly

Read each side of (A \cup B)' = A' \cap B' as a sentence about a random element x.

Both sides resolve to the identical sentence: "x is not in A and not in B." Same requirement, same set of x's. The law is a tautology disguised as an algebraic identity.

Why this works: the English word "either" in "not in either" implicitly means "neither A nor B." The implicit "neither" is a conjunction ("not in A and not in B"), which is exactly what A' \cap B' captures. De Morgan's law is telling you that "not in the union" and "in both complements" are two ways of saying "neither."

A household example

You are planning a dinner and you need to invite students who do not like spicy food and do not like raw onions. Let S be the set of spicy-food lovers and R be the set of raw-onion lovers. The invitees are S' \cap R' — the intersection of the two complements.

Now restate: you want students who do not like either spicy food or raw onions. That set is (S \cup R)' — the complement of the "likes at least one" group.

Both descriptions identify the same students. That identity is De Morgan's first law.

The Venn picture

Venn diagram proof of De Morgan's first lawTwo Venn diagrams side by side. On the left the region outside both circles is shaded, representing the complement of A union B. On the right the same region is shaded, representing A complement intersect B complement. The two shaded regions are identical. (A ∪ B)′ A B = A′ ∩ B′ A B
The shaded region on both diagrams is "everything outside the two circles." On the left, the single operation $(A \cup B)'$ produces it by flipping the interior of the merged blob. On the right, two complements combined with an intersection produce it by asking "outside $A$ *and* outside $B$." The eye sees immediately that the two shaded regions coincide.

The region outside both circles is characterised in two ways. Describe it as "not inside the merged region," and you have (A \cup B)'. Describe it as "outside A and outside B," and you have A' \cap B'. Same region; two different sentences name it.

Why does \cup turn into \cap?

This is the part that feels like a trick. When you push the complement inward, union flips to intersection. A common student reaction is "why not union stays union?"

Because the complement flips a logical "or" into a logical "and."

The jump "not (P or Q) = (not P) and (not Q)" is the logical version of the same De Morgan's law you learn in Logic and Propositions. The reason \cup becomes \cap is because "or" becomes "and" under negation. Set theory inherits this from logic directly.

Why "or" becomes "and": to fail a disjunction, you have to fail both disjuncts. A single success would make the "or" true. So the negation demands two simultaneous failures — and "simultaneous" is conjunction.

Trying the wrong version to feel the difference

What if someone wrote (A \cup B)' = A' \cup B' — keeping union on both sides? That would say "not in either circle equals not in A or not in B." Take x in the overlap A \cap B: it is in both circles, so the left side (A \cup B)' does not contain it. But x is either in A or it isn't — actually it is in A, so x \notin A', so x \in A' \cup B' iff x \in B'... wait, x \in B too, so x \notin B' either. So actually in the overlap case both sides agree on "does not contain x." That is not a counterexample.

Try x outside A but inside B. Then x \in A \cup B, so x \notin (A \cup B)' (left side: no). But x \notin A, so x \in A', so x \in A' \cup B' (right side: yes). Disagreement.

So the "wrong version" (A \cup B)' = A' \cup B' fails for any x that sits inside exactly one of the circles. The \cup \to \cap swap is not optional — without it, the identity breaks.

A numerical sanity check

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

Same result. The elements 6 and 7 are the only ones in neither A nor B, and both descriptions correctly identify them.

One-line proof via element-chasing

For any x \in U:

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

Four "if and only if" steps, each unpacking a definition. Every intermediate statement is the same logical sentence in slightly different dress. This element-chasing argument is the standard way to prove set identities — and it is the formal version of the plain-English sentence at the top.

A use-case in probability

Suppose you want P(\text{neither rain nor traffic}) tomorrow. Let R be the "rain" event and T the "traffic" event. The desired event is (R \cup T)'. By De Morgan, it equals R' \cap T' — the event "no rain and no traffic." Switching perspective gives:

P(\text{neither}) = 1 - P(R \cup T) \quad \text{or equivalently} \quad P(\text{no rain}) \cdot P(\text{no traffic} \mid \text{no rain}).

Both avenues reach the same number. De Morgan's law is the reason you can choose whichever is easier to compute.

The takeaway

You do not need to memorise "complement of union equals intersection of complements." You need to memorise the sentence:

Failing to be in either means failing to be in both.

Once that sentence is internal, the symbols (A \cup B)' = A' \cap B' become a transcription of something you already believe.

Related: Set Operations · De Morgan's Law Animation · Logic and Propositions · A − B vs A ∩ B′