Let A⊂ B . Simplify the expressions: 1^(°) AB; 2^(°) A+B…
Let $A\subset B$. Simplify the expressions: \[ 1^{\circ}\ AB; \qquad 2^{\circ}\ A+B; \qquad 3^{\circ}\ ABC; \qquad 4^{\circ}\ A+B+C. \]
Explanation of Method
$A\subset B$ means that whenever $A$ occurs, $B$ occurs too. So in a product, the factor $B$ adds no new requirement once $A$ is there, and in a sum, the term $A$ adds no new possibility once $B$ is there. Each simplification is proved by showing two inclusions.
Step 1: Part $1^{\circ}$ — $AB$
- $AB\subseteq A$: always true (if both occur, then $A$ occurs).
- $A\subseteq AB$: if $\omega\in A$, then $\omega\in B$ (because $A\subset B$), so $\omega\in AB$.
\[ \boxed{AB = A.} \]
In words: "$A$ and $B$" is just "$A$", since $B$ comes automatically with $A$.
Step 2: Part $2^{\circ}$ — $A+B$
- $B\subseteq A+B$: always true.
- $A+B\subseteq B$: if $\omega\in A+B$, then $\omega\in B$ directly, or $\omega\in A\subset B$.
\[ \boxed{A + B = B.} \]
In words: "$A$ or $B$" is just "$B$", since every time $A$ occurs, $B$ occurs anyway.
Step 3: Part $3^{\circ}$ — $ABC$
Group the first two factors and use part $1^{\circ}$:
\[ ABC = (AB)C = AC. \]
\[ \boxed{ABC = AC.} \]
In words: "all three occur" reduces to "$A$ and $C$ occur", because $B$ is implied by $A$. Note that in general $AC$ cannot be simplified further — nothing is known about $C$.
Step 4: Part $4^{\circ}$ — $A+B+C$
Group the first two terms and use part $2^{\circ}$:
\[ A+B+C = (A+B)+C = B+C. \]
\[ \boxed{A+B+C = B+C.} \]
In words: "at least one of $A,B,C$" reduces to "at least one of $B,C$", because whenever $A$ occurs, $B$ also occurs.
Step 5: Example with a die
$\Omega=\{1,\dots,6\}$, $A=\{2\}$, $B=\{2,4,6\}$ (so $A\subset B$), $C=\{1,2,3\}$.
| Expression | Direct computation | Simplified form | Match? |
|---|---|---|---|
| $AB$ | $\{2\}\cap\{2,4,6\}=\{2\}$ | $A=\{2\}$ | ✓ |
| $A+B$ | $\{2\}\cup\{2,4,6\}=\{2,4,6\}$ | $B=\{2,4,6\}$ | ✓ |
| $ABC$ | $\{2\}\cap\{2,4,6\}\cap\{1,2,3\}=\{2\}$ | $AC=\{2\}$ | ✓ |
| $A+B+C$ | $\{1,2,3,4,6\}$ | $B+C=\{1,2,3,4,6\}$ | ✓ |
Step 6: The general rule
If $A\subseteq B$, the "smaller" event $A$ may replace $B$ in a product, and the "larger" event $B$ may replace $A$ in a sum:
\[ AB=A, \qquad A+B=B. \]
(See also Problem 21: $A\subseteq B \iff A+B=B$.)
Result
| Expression | Simplified (given $A\subset B$) |
|---|---|
| $1^{\circ}\ AB$ | $A$ |
| $2^{\circ}\ A+B$ | $B$ |
| $3^{\circ}\ ABC$ | $AC$ |
| $4^{\circ}\ A+B+C$ | $B+C$ |