Visual Tools
Calculators
Tables
Mathematical Keyboard
Converters
Other Tools


Venn Diagrams: Three Sets Basic Identities


Regions of interest in three-set algebra

Jump to
DiagramA
UABC
Theme
ColorOpacity0.85
1 / 40
Explanation

Set A

A

Definition

The set A.

Example

A = {1,2,3}







Key Terms


  • Set — a collection of distinct elements
  • Universal set — the set containing every element under consideration, denoted UU
  • UnionABCA \cup B \cup C, elements in at least one of AA, BB, CC
  • IntersectionABCA \cap B \cap C, elements in all three sets
  • Pairwise intersectionABA \cap B, ACA \cap C, or BCB \cap C
  • ComplementAA', elements in UU but not in AA
  • Set differenceABA \setminus B, elements in AA but not in BB
  • Symmetric differenceABCA \triangle B \triangle C, elements in an odd number of AA, BB, CC
  • De Morgan's laws (three sets)(ABC)=ABC(A \cup B \cup C)' = A' \cap B' \cap C' and (ABC)=ABC(A \cap B \cap C)' = A' \cup B' \cup C'
  • Region — one of the eight disjoint pieces a three-circle Venn diagram divides the universe into


Getting Started with the Explorer

Open the explorer and a three-circle Venn diagram appears with the first identity pre-selected. The blue shaded regions mark the elements that satisfy the current identity; the unshaded regions are excluded. The symbol of the current identity appears in the badge above the diagram, and the explanation panel beside it describes what the highlighted regions mean.

The interface has three main controls. The category tabs at the top group identities by type. The formula buttons below the tabs show the identities within the active category. The Jump to dropdown on the right lists every identity across all categories in one place.

At the bottom of the diagram column, Previous and Next cycle through all 40 identities in order, with a counter showing your current position. The theme panel underneath lets you customize the shading color and opacity.

No setup is required — pick any tab and any button to see the corresponding region combination light up immediately.

Selecting an Identity

Two ways to pick an identity. Use the formula buttons under the active tab to choose from identities in that category — each button shows the set-theory notation (ABCA \cup B \cup C, AA', ABCA \triangle B \triangle C, "exactly two", and so on). Or use the Jump to dropdown, which lists every identity across all six categories in one menu, grouped by tab.

When you select an identity, three things update simultaneously:

• The diagram shading changes to highlight the regions belonging to the new identity
• The badge above the diagram updates to the new symbol
• The explanation panel refreshes with a definition and, where applicable, a numerical example like A={1,2,3}A = \{1,2,3\}, B={2,3,4}B = \{2,3,4\}, C={3,4,5}C = \{3,4,5\} for the triple intersection

The active category tab follows your selection, so you always see which group the current identity belongs to.

Reading the Shaded Venn Diagram

A three-circle Venn diagram divides the universe into eight disjoint regions, and every three-set identity highlights some combination of them:

Outside all three circles — elements in none of AA, BB, CC, formally ABCA' \cap B' \cap C'
Only in A — formally A(BC)A \setminus (B \cup C)
Only in B — formally B(AC)B \setminus (A \cup C)
Only in C — formally C(AB)C \setminus (A \cup B)
In A and B, not C — formally ABCA \cap B \cap C'
In A and C, not B — formally ABCA \cap B' \cap C
In B and C, not A — formally ABCA' \cap B \cap C
In all three — the triple intersection ABCA \cap B \cap C, the central region

For example, ABCA \cup B \cup C shades all seven regions inside any circle. "Exactly one" shades only the three "only" regions. "Exactly two" shades only the three pairwise-but-not-triple regions. (ABC)(A \cup B \cup C)' shades only the outside region. Hover over any region for a tooltip naming it.

Customizing Color and Opacity

The Theme panel below the diagram offers two adjustments to the shaded regions.

The color picker changes the shading hue. Useful when printing, presenting, comparing diagrams side by side, or matching the color scheme of a course or textbook. Any standard color value works.

The opacity slider controls how transparent the shading is, ranging from 1.001.00 (fully opaque) to 0.000.00 (invisible). Lower opacity is helpful when you want to see the underlying circle outlines through the fill, especially in the central triple-intersection region where multiple overlapping boundaries meet. The current numeric value appears next to the slider in monospace.

Click Reset to return both controls to the defaults — blue at 0.850.85 opacity. Theme changes persist as you navigate between scenarios, so adjustments stay applied across the entire session.

Previous and Next Navigation

At the bottom of the diagram column, the Previous and Next buttons cycle through all 40 identities in the order defined by the category groups: Basic Sets, then Complements, then Intersection & Union, then Differences, Compound, and finally De Morgan's Laws. The counter between the two buttons displays the current position, formatted as "nn / 4040".

Navigation wraps around: pressing Previous on the first scenario jumps to the last, and pressing Next on the last returns to the first. This makes the explorer well suited for systematic review — start at the first identity and click through every region combination one by one to see how each algebraic expression maps to a subset of the eight regions.

The active tab and active formula button update automatically as you advance, so you always know which group the current identity belongs to.

The Basic Sets

The Basic Sets tab shows the diagram's raw material before any operation acts on it. Set A, set B, and set C each shade one full circle — which in a three-circle diagram already means four regions apiece. The universal set shades all eight regions, and the empty set shades none.

The step up from two sets is worth pausing on: each circle now overlaps *two* neighbors plus the shared center, so even "just the set AA" is a four-region composite. Learning to see one circle as four regions is the core reading skill for everything below.

The Set A on Its Own

The top circle, fully shaded — and in a three-set diagram that already means four regions: A-only, the two pairwise slivers ABA \cap B and ACA \cap C, and the central core.
UABC
Set A, frozen

The full top circle: four regions at once — one private, two shared slivers, one center. A single set is already a composite here.

The four-piece decomposition A=(A only)(ABC)(ABC)(ABC)A = (A \text{ only}) \cup (A \cap B \cap C') \cup (A \cap B' \cap C) \cup (A \cap B \cap C) is the three-set version of the two-piece split familiar from two circles, and it drives every counting argument about A|A| in the presence of two other sets.

With the example A={1,2,3}A = \{1,2,3\}, the four pieces sort the elements by which neighbors also claim them. Compare set B and set C — the same frame rotated a third of a turn.

The Set B on Its Own

The lower-left circle, fully shaded: B-only, the slivers shared with AA and with CC, and the center.
UABC
Set B, frozen

The lower-left circle, fully lit. Rotate the frame a third of a turn and it becomes set A — commutativity as rotational symmetry.

Nothing distinguishes BB's frame from $A$'s except position — rotate the diagram 120°120° and they trade places. That rotational symmetry is the three-set face of commutativity: any identity indifferent to the naming of sets must produce rotation-symmetric families of frames.

The decomposition reads the same way: BB splits into its private region plus three shared pieces, and the explorer's tooltips name each one on hover.

The Set C on Its Own

The lower-right circle, fully shaded — the third and last rotation of the single-set frame.
UABC
Set C, frozen

The third rotation completes the family: every single-set frame is one private region plus three co-owned pieces.

With all three single-set frames seen, a bookkeeping fact becomes visible: each of the seven inside regions is claimed by one, two, or three of the circles, and summing A+B+C|A| + |B| + |C| counts the pairwise slivers twice and the center three times. That over-counting is exactly what the inclusion-exclusion formula corrects.

The frame also completes the reading drill: whichever circle is shaded, the picture is four regions — one private, two shared with a single neighbor, one shared with both. See only in C for the private region alone.

The Universal Set U

All eight regions shaded — the three-set diagram at full saturation.
UABC
Universal set U, frozen

All eight regions shaded — the 2³ membership patterns of three sets, saturated at once.

The rectangle's role is unchanged from the two-set case: UU is the modeling boundary, identity for intersection, annihilator for union, and the reference against which the three complements are cut. What changes is only the count — the universe now decomposes into 23=82^3 = 8 disjoint regions, one for each membership pattern.

Its polar opposite is the empty set; the pair bound the whole catalog between "everything" and "nothing".

The Empty Set

Zero regions shaded: three interlocking circles and no highlight anywhere.
UABC
Empty set ∅, frozen

Three interlocking circles, zero highlight. The blank frame is the picture of expressions like A ∩ A′.

As with two sets, the blank frame is a real answer — AAA \cap A', or the intersection of any disjoint pair, produces exactly this shading. And \emptyset remains a subset of every set, vacuously.

The blank frame is also the catalog's zero point: every other state is "the empty set plus some regions". Clicking between \emptyset and the universal set shows the two extremes that complementation, in the complements tab, exchanges.

The Three Complements

The Complements tab flips one circle at a time: the complement of A, the complement of B, and the complement of C. Each shades four regions — the outside plus everything that avoids the complemented circle.

The three frames are rotations of one another, a first taste of the diagram's three-fold symmetry: any statement proved about AA' transfers to BB' and CC' by relabeling. Complements are also the raw ingredients of the De Morgan states at the end of the catalog.

The Complement of A

Four regions shaded: the outside, B-only, C-only, and the BCB \cap C sliver — everything that avoids the top circle.
UABC
Complement A′, frozen

Four regions avoiding the top circle: outside, both neighbors’ private parts, and their shared bottom sliver.

The complement of one set among three is bigger than beginners expect: it keeps most of BB and most of CC. Only the four regions inside circle AA — including the slivers ABA \cap B and ACA \cap C and the center — are excluded, because their elements are in AA regardless of what else they belong to.

Together with set A on its own, the two frames partition all eight regions four-and-four: the complement laws AA=UA \cup A' = U, AA=A \cap A' = \emptyset survive the move to three sets untouched.

The Complement of B

The rotation of the previous frame: outside, A-only, C-only, and the ACA \cap C sliver — the four regions avoiding the lower-left circle.
UABC
Complement B′, frozen

The rotation: only the A∩C sliver survives among the shared pieces — the one overlap with no stake in B.

Note which shared piece survives: ACA \cap C is the one pairwise sliver with no stake in BB, so it is the only two-set region inside BB'. The center never survives any complement — its elements are in all three sets, so every complement evicts them.

Rotating once more gives the complement of C; rotating back, the complement of A. All three are one frame seen from three angles.

The Complement of C

The third rotation: outside, A-only, B-only, and the ABA \cap B sliver.
UABC
Complement C′, frozen

Third rotation. The center never survives any complement: its elements belong to all three sets.

With all three complements seen, an intersection experiment suggests itself: overlay any two of them and only two regions survive both — the outside and the third set's private region. Overlay all three and only the outside remains, which is precisely the complement of the triple union, the first De Morgan state.

That mental overlay — intersecting complements region by region — is the three-set De Morgan law being computed by eye.

Intersections and Unions

Eight states cover the and/or combinations. At the extremes sit the triple intersection — the diagram's smallest interesting region — and the triple union, everything inside any circle. Between them, the three pairwise intersections (A ∩ B, A ∩ C, B ∩ C) and the three pairwise unions (A ∪ B, A ∪ C, B ∪ C).

The pairwise states carry the tab's key lesson: with a third set on the page, ABA \cap B is *two* regions and ABA \cup B is *six*, because the ignored set CC cuts through both. An operation on two sets does not stop being about two sets — but its picture changes when the universe holds a third.

The Triple Intersection

A single region shaded: the central core where all three circles overlap, ABCA \cap B \cap C.
UABC
A ∩ B ∩ C, frozen

One region, the central core — the only address that passes all three membership tests. Example sets leave just {3} here.

The center is the diagram's most exclusive address — membership requires passing all three tests. With the example sets A={1,2,3}A = \{1,2,3\}, B={2,3,4}B = \{2,3,4\}, C={3,4,5}C = \{3,4,5\}, only the element 33 qualifies.

Structurally, the triple intersection is inside every pairwise intersection, which is why each pairwise state — see A ∩ B — shades the center along with its own sliver. It is also the region that separates "exactly two" from "at least two" in the counting states.

The Pairwise Intersection A ∩ B

Two regions: the ABA \cap B sliver plus the central core. "In both AA and BB" says nothing about CC — so both answers to the CC question are included.
UABC
A ∩ B, frozen

Two regions: the A∩B sliver plus the center. “In both A and B” leaves the C question open — so both answers are in.

This is the tab's recurring subtlety. The region labeled ABA \cap B in the diagram is really ABCA \cap B \cap C' — the *pairwise-but-not-triple* piece — while the full set ABA \cap B is that sliver together with the center. The explorer's shading draws the distinction sharply.

Want the sliver alone? That is a different expression, ABCA \cap B \cap C', frozen at A and B but not C in the Compound tab.

The Pairwise Intersection A ∩ C

The rotation: the ACA \cap C sliver plus the center — everything in both AA and CC, with BB left unexamined.
UABC
A ∩ C, frozen

The rotated pairwise intersection: its own sliver plus the shared center.

The same two-region anatomy as A ∩ B, reflected to the other side of the top circle. Each pairwise intersection contains the triple intersection, so the three states of this family share the center and differ only in which sliver joins it.

Its not-BB restriction lives at A and C but not B.

The Pairwise Intersection B ∩ C

The third rotation: the bottom sliver BCB \cap C plus the center — the pairwise intersection that ignores AA.
UABC
B ∩ C, frozen

The bottom pairwise intersection — third of the family, same two-region anatomy.

Completing the family shows the pattern: three pairwise intersections, three slivers, one shared center. Their union is exactly the "at least two" counting state, and removing the center from each gives the "exactly two" state — the counting quartet is assembled from these parts.

The corresponding restricted region is B and C but not A.

The Triple Union

Seven regions shaded — everything inside any circle. Only the outside stays blank.
UABC
A ∪ B ∪ C, frozen

Seven of the eight regions — everything inside any circle. The disjoint pieces inclusion-exclusion adds and subtracts.

The triple union is the counting workhorse of three-set problems. Its seven disjoint regions are what the three-set inclusion-exclusion formula tallies: add the three sets, subtract the three pairwise intersections (each counted twice), and add back the triple intersection (counted three times, subtracted three times):

ABC=A+B+CABACBC+ABC|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|


Its photographic negative — the one blank region here — is the complement of the triple union.

The Pairwise Union A ∪ B

Six regions shaded: everything except the outside and C-only. A two-set union — drawn inside a three-set universe.
UABC
A ∪ B, frozen

Six regions, not three: C slices through both circles, so its slivers and the center ride along. Only C-only and the outside are spared.

Why six regions and not three? Because CC slices through both circles: the slivers ACA \cap C, BCB \cap C, and the center all contain elements of AA or BB, so they belong to the union even though they also touch CC. The only inside region excluded is C-only — in CC and in nothing else.

Compare the rotations A ∪ C and B ∪ C: each pairwise union excludes exactly the third set's private region.

The Pairwise Union A ∪ C

Six regions again — this time only the outside and B-only are spared.
UABC
A ∪ C, frozen

The rotation spares B-only instead — a pairwise union always excludes exactly the third set’s private region.

The frame confirms the family rule seen in A ∪ B: a pairwise union in a three-set universe misses exactly two regions, the outside and the ignored set's private piece. Everything else has at least one foot in AA or CC.

The complement of this shading — B-only plus outside — is a compound expression worth writing down as an exercise: (AC)=AC(A \cup C)' = A' \cap C', the two-set De Morgan law operating inside the three-set picture.

The Pairwise Union B ∪ C

The final rotation: six shaded regions, sparing the outside and A-only.
UABC
B ∪ C, frozen

The final rotation: six regions, A-only left out. This is the “B or C” that later gets clipped by A in the compound tab.

This union is the "BB or CC" that appears inside the compound state A ∩ (B ∪ C) — freezing it separately makes the later composition readable: first form the six-region union, then intersect with the top circle.

Excluding A-only is also a way to see subtraction hiding in union: the shaded set is U(A(BC))U \setminus (A \setminus (B \cup C))'... which is precisely why the explorer prefers pictures to nested formulas.

The Differences

Twelve subtraction states. The six one-sided differences (A ∖ B, A ∖ C, B ∖ A, B ∖ C, C ∖ A, C ∖ B) each shade two regions. Subtracting *both* neighbors instead of one produces the three "only" states (only in A, only in B, only in C). The tab closes with the mixed difference (A ∪ B) ∖ C and the two symmetric differences — A △ B and the triple form A △ B △ C.

The organizing contrast: ABA \setminus B keeps the part of AA that CC shares, while A(BC)A \setminus (B \cup C) strips AA to its private region. One extra subtraction, one region fewer — the difference between "not in BB" and "in nothing else".

The Difference A Minus B

Two regions: A-only plus the ACA \cap C sliver — the part of the top circle that stays clear of BB.
UABC
A ∖ B, frozen

Two regions: A-only plus the A∩C sliver. Subtracting B removes B’s sliver and the center — the C overlap is an innocent bystander.

The three-set surprise is the second region: ABA \setminus B keeps the elements AA shares with CC, because the subtraction only asks about BB. Removing a set removes its slivers and the center, but the "innocent bystander" overlap with CC survives.

To strip that too, subtract both neighbors: only in A shades the truly private region. The one-sided differences and the only-regions differ by exactly one sliver each.

The Difference A Minus C

The mirror subtraction: A-only plus the ABA \cap B sliver — the top circle cleared of CC.
UABC
A ∖ C, frozen

The mirror subtraction keeps the other sliver: same minuend, different subtrahend, different survivor.

Together with A minus B, this frame shows subtraction's direction-sensitivity inside one circle: same minuend AA, different subtrahend, different surviving sliver. Both frames keep A-only; they disagree only about which neighbor's overlap lives.

The rewrite AC=ACA \setminus C = A \cap C' still holds — every difference in this tab is an intersection with a complement in disguise.

The Difference B Minus A

Two regions in the lower left: B-only plus the bottom sliver BCB \cap C.
UABC
B ∖ A, frozen

Reversal moves the shading to the other circle entirely — the two differences of a pair share nothing.

Reversing A minus B moves the shading to the other circle entirely — the two one-sided differences of a pair share no region at all, exactly as in the two-set case, only now each is a two-region composite.

The surviving sliver is again the bystander: BB's overlap with CC has no stake in the subtraction of AA.

The Difference B Minus C

B-only plus the ABA \cap B sliver: the lower-left circle with everything touching CC removed.
UABC
B ∖ C, frozen

B-only plus the A∩B sliver: the mechanical pattern — delete the subtrahend’s sliver and the center, keep the rest.

The pattern by now is mechanical: subtracting one neighbor deletes that neighbor's sliver and the center, keeps the private region and the other sliver. Six one-sided differences, six two-region frames, each a rotation or reflection of the others.

What stays constant across the family is the rewrite XY=XYX \setminus Y = X \cap Y' — worth checking mentally against the frame each time.

The Difference C Minus A

C-only plus the bottom sliver BCB \cap C: the lower-right circle cleared of AA.
UABC
C ∖ A, frozen

C-only plus the bottom sliver. Note it shares that sliver with B ∖ A — both subtractions of A spare it.

A useful comparison: B minus A and this frame share the BCB \cap C sliver — both subtractions of AA keep it, since its elements avoid AA by definition. Overlaying the two frames would shade (BC)A(B \cup C) \setminus A exactly.

That overlay observation is a difference identity being discovered pictorially: (BA)(CA)=(BC)A(B \setminus A) \cup (C \setminus A) = (B \cup C) \setminus A.

The Difference C Minus B

The last of the six: C-only plus the ACA \cap C sliver.
UABC
C ∖ B, frozen

The last one-sided difference. None of the six ever shade the center — subtracting any set evicts it.

Closing the one-sided family, note what none of the six frames ever shade: the center. The triple intersection belongs to every set, so subtracting any one set always evicts it — only the "only" regions and single slivers survive one-sided subtraction.

The stricter cut that removes the surviving sliver too is only in C, one step down the tab.

Only in A

One region: the top of the diagram, A(BC)A \setminus (B \cup C) — elements of AA and of nothing else.
UABC
A ∖ (B ∪ C), frozen

One region: the truly private part of A. Subtracting the union of both neighbors is the strongest single-set exclusion.

Subtracting the union of both neighbors is the strongest exclusion the diagram offers a single set. Compare A minus B: one extra set in the subtrahend costs exactly one more region, the ACA \cap C sliver.

The three "only" regions are the atoms of the counting states — exactly one is precisely their union — and in survey problems they answer the question "how many chose *only* this option?"

Only in B

The private region of the lower-left circle: B(AC)B \setminus (A \cup C), one region, no shared pieces.
UABC
B ∖ (A ∪ C), frozen

“Not in A or C” = “not in A and not in C” — De Morgan quietly at work in the formula for one quiet region.

The formula deserves one careful reading: subtracting ACA \cup C is subtracting *either* neighbor, so the survivors avoid both. This is De Morgan operating quietly — "not in AA or CC" equals "not in AA and not in CC", i.e. BACB \cap A' \cap C'.

The three only-frames are the diagram's rotational family at its purest: one region each, at three clock positions.

Only in C

The lower-right private region: C(AB)C \setminus (A \cup B).
UABC
C ∖ (A ∪ B), frozen

The third private region. The three “only” frames are the atoms of the counting states.

Completing the trio makes the "exactly one" decomposition concrete: the three private regions are pairwise disjoint, so exactly one=A only+B only+C only|{\text{exactly one}}| = |A \text{ only}| + |B \text{ only}| + |C \text{ only}| with no correction terms — disjointness is what makes counting additive.

Each only-region is also what remains of its set in the strictest sense; everything else a set owns is co-owned. The contrast between ownership and co-ownership is the whole story of the differences tab.

The Union of A and B Minus C

Three regions: A-only, B-only, and their shared sliver ABA \cap B — the two-circle union with everything touching CC carved away.
UABC
(A ∪ B) ∖ C, frozen

Three regions: the two-circle union with every C-touching piece carved away — a two-set world restored at the top of the frame.

The frame is best read as a two-step: the pairwise union A ∪ B shades six regions; subtracting CC then removes the three that touch the lower-right circle (ACA \cap C, BCB \cap C, and the center). What remains is a self-contained two-set world — the top of the diagram behaving as if CC did not exist.

Mixed expressions like this one are the bridge between the difference and compound tabs: one more grouping variant and the distributive laws come into view.

The Symmetric Difference of A and B

Four regions: A-only, B-only, and the two side slivers ACA \cap C and BCB \cap C — in exactly one of AA or BB, with CC free to do as it pleases.
UABC
A △ B, frozen

Four regions: each two-set crescent is split by C into a private piece and a sliver. Exactly one of A, B — C rides free.

The two-set symmetric difference imported into a three-set universe keeps its meaning — membership in exactly one of AA, BB — but its picture doubles: each "crescent" of the two-set diagram is now split by CC into a private piece and a sliver. The excluded regions are those where the AA/BB count is 00 or 22: the outside, C-only, the ABA \cap B sliver, and the center.

The genuinely three-set generalization is the triple symmetric difference, where the parity rule takes over.

The Triple Symmetric Difference

Four regions in the tab's most surprising pattern: the three "only" regions — plus the center.
UABC
A △ B △ C, frozen

The parity rule: an odd membership count. The three only-regions (count 1) plus the lit center (count 3).

ABCA \triangle B \triangle C collects elements in an *odd number* of the three sets: one or three. The only-regions supply the count-one elements; the triple intersection supplies count-three. The pairwise slivers, at count two, are excluded — which is why the shading skips them and the frame looks like "exactly one" with a lit center.

The parity rule is what makes \triangle associative: however the expression is parenthesized, an element's membership flips once per set that contains it, and only an odd number of flips leaves it in. Compare exactly one — identical but for the center.

Compound Expressions

The Compound tab is where three sets earn their keep. Three states isolate single pairwise regions (A and B but not C, A and C but not B, B and C but not A); three show mixed and/or grouping (A ∩ (B ∪ C), (A ∪ B) ∩ C, A ∪ (B ∩ C)); and four count memberships (exactly one, exactly two, at least two, at most one).

The counting quartet has no two-set analogue, and the grouped states preview the distributive laws: A(BC)A \cap (B \cup C) and A(BC)A \cup (B \cap C) shade different region sets, proving parentheses matter when the operations mix.

In A and B but Not C

A single region: the ABA \cap B sliver alone, ABCA \cap B \cap C' — the pairwise overlap with the center explicitly evicted.
UABC
A ∩ B ∩ C′, frozen

The A∩B sliver alone — the pairwise overlap with the center explicitly evicted. Three literals, one region.

This is the state that resolves the pairwise-intersection subtlety head-on: the full A ∩ B is two regions, and appending C\cap\, C' deletes the center, leaving the sliver. Three symbols in the formula, one region in the picture — each of the eight regions is expressible as such a triple condition.

In survey language: "chose AA and BB but not CC" — the kind of clause inclusion-exclusion bookkeeping constantly needs.

In A and C but Not B

The rotation: the ACA \cap C sliver alone, ABCA \cap B' \cap C.
UABC
A ∩ B′ ∩ C, frozen

The rotated sliver: every region of the diagram is one such three-literal conjunction. Eight combinations, eight regions.

Same anatomy as A and B but not C, one position around the ring. The three sliver-states are the middle tier of the membership hierarchy — more exclusive than a full set, less exclusive than the center.

Note the formula's shape: every region of the diagram is a conjunction of three literals, one per set, primed or unprimed. Eight combinations, eight regions — the diagram *is* the truth table of three variables.

In B and C but Not A

The bottom sliver alone: ABCA' \cap B \cap C.
UABC
A′ ∩ B ∩ C, frozen

The bottom sliver alone — third atom of “exactly two”.

The third sliver completes the "exactly two" atoms: unite the three sliver-states and you get exactly two of the three sets region for region. The compound tab first shows the parts, then the assemblies.

The primed-first formula (AA' leading) reads oddly aloud — "not in AA, in BB, in CC" — but the alphabetical discipline pays off when comparing all eight region-formulas side by side.

A Intersected with B Union C

Three regions: both of AA's slivers plus the center — the part of the top circle that touches at least one neighbor.
UABC
A ∩ (B ∪ C), frozen

The social half of circle A: both slivers plus the center. Left side of a distributive law, drawn.

Read it inside-out: B ∪ C shades six regions, and intersecting with AA keeps only those inside the top circle. Equivalently, subtract this from full AA and what is left is only in A — this state and the only-state split the circle into social and solitary halves.

The frame is also the left side of a distributive law: A(BC)=(AB)(AC)A \cap (B \cup C) = (A \cap B) \cup (A \cap C) — and the right side visibly assembles the same three regions from the two pairwise intersections.

A Union B Intersected with C

Three regions again, rotated onto the lower-right circle: CC's two slivers plus the center — the part of CC touching AA or BB.
UABC
(A ∪ B) ∩ C, frozen

The rotated twin: a two-circle union clipped to the third circle — always the clipped circle minus its private region.

Structurally the twin of A ∩ (B ∪ C) with the roles rotated: form the union of two circles, then clip to the third. The result is always the clipped circle minus its private region.

Comparing the two frames drives home that the *pattern* of a compound expression, not the letters in it, determines the picture's shape — relabeling rotates the frame, regrouping changes it.

A United with B Intersect C

Five regions: all four pieces of circle AA, plus the bottom sliver BCB \cap C — the whole top circle joined by the one region where BB and CC meet without it.
UABC
A ∪ (B ∩ C), frozen

Five regions: all of A plus the bottom sliver. Swap ∩ and ∪ in the grouping and the picture inflates — parentheses matter.

Swap the operations of A ∩ (B ∪ C) and the picture inflates from three regions to five: union is generous where intersection is strict. Setting the two frames side by side is the fastest proof that \cap and \cup do not commute with each other — grouping matters.

This frame is likewise half of a distributive law: A(BC)=(AB)(AC)A \cup (B \cap C) = (A \cup B) \cap (A \cup C), whose right side intersects two six-region unions down to these same five regions.

Exactly One of the Three Sets

The three "only" regions — membership count equal to one, nothing shared, nothing outside.
UABC
Exactly one, frozen

The three private regions — membership count exactly 1. A counting state with no real two-set ancestor.

The first of the counting quartet, and the one with no two-set ancestor worth the name. Its atoms were built in the differences tab (only in A and its rotations); here they are united into a single predicate on membership count.

Counting states are where the diagram meets applications: "exactly one" answers survey questions ("chose a single option"), probability questions (exactly one event occurs), and — with the center added — becomes the triple symmetric difference.

Exactly Two of the Three Sets

The three pairwise slivers — membership count exactly two, the center pointedly excluded.
UABC
Exactly two, frozen

The three pairwise slivers, center excluded — count exactly 2. The picture is far more legible than its formula.

The quartet's middle state assembles the three sliver-atoms from the compound singles. Its formula, written out, is a three-way union of triple conjunctions — the picture is far more legible than the algebra, which is the explorer's argument in miniature.

Add the center and the count relaxes to at least two; the difference between the two states is one region and one word.

At Least Two of the Three Sets

Four regions: the three pairwise slivers plus the center — membership count two or three.
UABC
At least two, frozen

Slivers plus center — the majority rule, two-out-of-three. The whole interior overlap structure at once.

"At least two" is the majority state: its elements are in most of the sets. It decomposes as exactly two plus the triple intersection, and the frame shows the join seamlessly — the four regions form the diagram's whole interior overlap structure.

In voting terms this is the two-out-of-three rule; in probability, the event "at least two of three events occur" — the direct three-set generalization of an intersection.

At Most One of the Three Sets

Four regions at the diagram's periphery: the outside plus the three "only" regions — membership count zero or one.
UABC
At most one, frozen

The periphery: outside plus the three private regions — count 0 or 1. The photographic negative of “at least two”.

The quartet closes with the complement of the previous state: "at most one" and at least two partition the eight regions four-and-four, since every element's membership count is either ≤ 1 or ≥ 2. The two frames are photographic negatives.

It is also the only counting state that shades the outside — count zero qualifies — a reminder that membership predicates quantify over all of UU, not just the circles.

What is a Three-Set Venn Diagram?

A three-set Venn diagram is a visual representation of three sets drawn as three overlapping circles inside a rectangle. The rectangle represents the universal set UU — everything under consideration. The three circles, labeled AA, BB, and CC, are arranged symmetrically so that every possible combination of memberships produces its own region.

The diagram has exactly eight disjoint regions: one outside all circles, three "only" regions (only in AA, only in BB, only in CC), three pairwise-but-not-triple regions, and the central triple intersection. Every algebraic combination of three sets — no matter how complex — maps to some union of these eight regions.

This is what gives three-set diagrams their distinctive value: they can visualize "counting" identities like "exactly two of A,B,CA, B, C" or "at least one of A,B,CA, B, C" that have no analogue in the two-set case, where there is no notion of "exactly two of two."

For comprehensive theory on Venn diagrams across different numbers of sets, see Venn diagrams.

Three-Set Operations

Three-set algebra is generated by the same five operations as two-set algebra, but applied to three operands:

Triple union ABCA \cup B \cup C — elements in at least one of AA, BB, CC. Visually, every region inside any circle.

Triple intersection ABCA \cap B \cap C — elements in all three sets simultaneously. Visually, the central region where all three circles overlap.

Pairwise intersection ABA \cap B — elements in both AA and BB, regardless of CC. Visually, two regions: the "in A and B, not C" region and the central triple intersection.

Complement AA' — elements outside AA. With three sets this shades four regions: outside all circles, only in BB, only in CC, and the BCB \cap C region.

Difference ABA \setminus B — elements in AA but not in BB. Two regions: "only in A" and "in A and C, not B".

Symmetric difference ABCA \triangle B \triangle C — elements in an odd number of the three sets. Four regions: the three "only" regions plus the triple intersection.

For formal definitions and algebraic properties, see set operations.

De Morgan's Laws for Three Sets

De Morgan's laws extend cleanly from two sets to three:

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


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


The complement of the triple union equals the intersection of the three complements. The complement of the triple intersection equals the union of the three complements. The pattern generalizes to any finite collection of sets.

Both laws can be verified visually with the explorer. Select (ABC)(A \cup B \cup C)' from the De Morgan's Laws tab: only the region outside all three circles is shaded. That same region is what ABCA' \cap B' \cap C' would produce — outside AA and outside BB and outside CC simultaneously. Likewise, (ABC)(A \cap B \cap C)' shades everything except the central triple intersection — the same regions that ABCA' \cup B' \cup C' produces.

For algebraic proofs, the general nn-set form, and applications to propositional logic, see De Morgan's laws.
Each law has a dedicated frozen frame below: the complement of the triple union — the catalog's one-region extreme — and the complement of the triple intersection, its seven-region mirror.

The Complement of the Triple Union

One region: the outside. Beyond the union of all three sets means beyond each of them individually.
UABC
(A ∪ B ∪ C)′, frozen

One region, the outside: beyond the union means beyond each set individually — the intersection of all three complements.

The three-set first De Morgan law, (ABC)=ABC(A \cup B \cup C)' = A' \cap B' \cap C', is this frame's caption: the lone shaded region is exactly what survives intersecting the three complements — each complement keeps the outside, and the outside is all they share.

It is the negative of the triple union: seven regions there, the eighth here. The generalization to nn sets changes nothing but the count.

The Complement of the Triple Intersection

Seven regions — everything except the central core. Avoiding "all three" merely requires missing one.
UABC
(A ∩ B ∩ C)′, frozen

Seven regions, only the core spared: escaping “all three” takes just one failed membership. The 1-versus-7 mirror of the union law.

The second law, (ABC)=ABC(A \cap B \cap C)' = A' \cup B' \cup C', shades the majority of the diagram: an element escapes the triple intersection by failing any single membership test, and only the center passes all three. One region spared here, one region shaded in the union law — the same 1-versus-7 mirror the two-set laws showed as 1-versus-3.

Ending the catalog on this pair is apt: the two frames compress everything the page teaches — complements, unions, intersections, and the eight-region anatomy — into a single visual contrast.

Counting Identities Unique to Three Sets

Three-set algebra is the smallest setting where "counting" identities become non-trivial. The Compound tab collects them:

Exactly one of $A, B, C$ — elements in exactly one of the three sets. Shades the three "only" regions.

Exactly two of $A, B, C$ — elements in exactly two of the three sets. Shades the three pairwise-but-not-triple regions, excluding the central triple intersection.

At least two of $A, B, C$ — elements in two or three sets. Combines "exactly two" with the triple intersection.

At most one of $A, B, C$ — elements in zero or one sets. Combines the outside region with the three "only" regions.

These identities are typical of how three-set Venn diagrams are applied in combinatorics, probability (inclusion-exclusion), and survey analysis. The two-set case collapses most of them — "exactly two of two" is just the intersection, "at least two of two" is also the intersection.

For comprehensive treatment, see set laws and identities.