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Venn diagrams

B

2
A 4 5 C

1 7 3
6
8
S

Figure 1.7 – The general Venn diagram for three events is divided into eight regions

PROOF. Show that 1. A ∪ (A ∩ B) = A ∩ (A ∪ B) = A, 2. (A − B) ∪ (A ∩ B) = A.
1. Using the distributivity and idempotency laws above, we see that

A ∪ (A ∩ B) = (A ∪ A) ∩ (A ∪ B) = A ∩ (A ∪ B).
By sketching a Venn diagram it is immediately clear that both expressions are equal to A. Nevertheless,
we here proceed in a more formal manner in order to deduce this result algebraically. Let us begin by
writing
X = A ∪ (A ∩ B) = A ∩ (A ∪ B), (1.15)
fromwhichwewanttodeduceasimplerexpressionfortheeventX.Usingthefirstequalityin(1.15)and
the algebraic laws for ∩ and ∪, we may write

A ∩ X = A ∩ [A ∪ (A ∩ B)] = (A ∩ A) ∪ [A ∩ (A ∩ B)] = A ∪ (A ∩ B) = X.
Since A ∩ X = X we must have X ⊂ A. Now, using the second equality in (1.15) in a similar way, we
find

A ∪ X = A ∪ [A ∩ (A ∪ B)] = (A ∪ A) ∩ [A ∪ (A ∪ B)] = A ∩ (A ∪ B) = X,
from which we deduce that A ⊂ X. Thus, since X ⊂ A and A ⊂ X, we must conclude that X = A.
2. Since we do not know how to deal with compound expressions containing a minus sign, we begin by
¯
writing A − B = A ∩ B mentioned above. Then, using the distributivity law, we obtain
¯
¯
(A − B) ∪ (A ∩ B) = (A ∩ B) ∪ (A ∩ B) = (A ∩ B) == A ∩ (B ∪ B) = A ∩ S = A.
In fact, this result, like the first one, can be proved trivially by drawing a Venn diagram. 2

Further useful results may be derived from Venn diagrams. In particular, it is simple to show that
the following rules hold:
1. if A ⊂ B then A ⊃ B;
2. A ∪ B = A ∩ B;
3. A ∩ B = A ∪ B.
Statements (2) and (3) are known jointly as de Morgan’s laws and are sometimes useful in
simplifying logical expressions.
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Example 1.12. There exist two events A and B such that (X ∪ A) ∪ (X ∪ A) = B.
Find an expression for the event X in terms of A and B. ,

Solution. We begin by taking the complement of both sides of the above expression: applying
de Morgan’s laws we obtain
¯
¯
B = (X ∪ A) ∩ (X ∪ A).
¯
¯
We may then use the algebraic laws obeyed by ∩ and ∪ to yield B = X ∪ (A ∩ A) = X ∪ ∅ = X.
¯
Thus, we find that X = B.
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