De Morgan's first law says (A \cup B)' = A' \cap B'. On paper it is a short algebraic identity; on a Venn diagram it is something better — two descriptions of exactly the same region. The trick is to watch one description morph into the other without the shaded area ever moving.

That is the whole point of this animation. You drag a slider from 0 to 1. At t=0 the picture highlights (A \cup B)' — the area outside both circles. At t=1 the picture highlights A' \cap B' — the overlap of "outside A" and "outside B." The shaded pixels are identical the whole way through. Only the story changes.

The identity, in one line

(A \cup B)' = A' \cap B'

Read it as an instruction for flipping a shaded region:

If you stare at either instruction, you land on the same picture: the part of the rectangle that lies outside both circles. The animation makes that equivalence visible.

The interactive picture

Slide the control below. The number t controls how much colour each description contributes. At t = 0 you see the "(A \cup B)'" story — a single flood filling the outside. At t = 1 you see the "A' \cap B'" story — two tinted strips (outside A, outside B) whose overlap lights up.

Slider-driven animation of De Morgan's first lawA rectangle contains two overlapping circles labelled A and B. A slider at the bottom controls a parameter t between 0 and 1. At t=0 the region outside both circles is shaded as the complement of the union. At t=1 the same region is shaded as the intersection of the complement of A and the complement of B. A live readout reports the current framing. U A B outside both circles (always shaded) drag slider from 0 to 1
The shaded outer region never changes. What changes is the description. For the first half of the slide, you see it as $(A \cup B)'$ — the flood outside a single merged blob. For the second half, you see it as $A' \cap B'$ — the overlap of "everywhere not in $A$" with "everywhere not in $B$." The shaded region is the same set of points.

Why the two stories agree: a point lies outside A \cup B exactly when it is outside both circles. "Outside the merged blob" and "outside A and also outside B" are word-for-word the same requirement.

A point-by-point check

Pick any point p in the rectangle. For De Morgan's first law to hold, these two statements must be true or false together:

Both statements say "p is not in A and not in B." Same requirement, same set. The animation is just a moving proof of this equivalence.

Four labelled points

Pick four specific locations and trace them through both descriptions.

Point In A? In B? In A \cup B? In (A \cup B)'? In A' \cap B'?
centre of A only yes no yes no no
centre of B only no yes yes no no
inside the overlap yes yes yes no no
outside both circles no no no yes yes

Only the fourth row is in the shaded region — and in that row, both of the last two columns read "yes." The identity is row-by-row.

The sister law

De Morgan's second law swaps union and intersection:

(A \cap B)' = A' \cup B'

The animation for this one would flood "outside the lens" at t=0 and "outside A or outside B" at t=1. Same game — both descriptions land on the same region.

You can remember the pair with one slogan: complement swaps the operation. Union becomes intersection, intersection becomes union, each time a complement bar is distributed inwards.

A classroom-sized example

Let U = \{1, 2, \dots, 10\}, A = \{1, 2, 3, 4\}, B = \{3, 4, 5, 6\}.

Same four elements. The identity works element by element, not just region by region.

Why: element 7 is in neither A nor B, so it passes the test for both sides. Elements 1, 2, 3, 4, 5, 6 are in at least one of the circles, so they fail both sides. The match is forced.

Where this matters

De Morgan's laws are not a curiosity — they are a daily tool once you start proving set identities or designing logic circuits.

Related: Set Operations · Animated Venn Diagrams · Symmetric Difference · Logic and Propositions