When are the following equalities possible: 1^(°)…
When are the following equalities possible: \[ 1^{\circ}\ A+B=\overline{A}; \qquad 2^{\circ}\ AB=\overline{A}; \qquad 3^{\circ}\ A+B=AB\ ? \]
Explanation of Method
We combine the equality with the facts that are true for all events: $A\subseteq A+B$, $AB\subseteq A$, and $AB\subseteq A+B$. This gives inclusions such as $A\subseteq\overline{A}$. An event can be contained in its own opposite only if it is impossible, because an outcome cannot both belong and not belong to $A$. Such inclusions quickly force the events to be $\varnothing$ (impossible) or $\Omega$ (certain). At the end we check that the conditions found really make the equality true.
Step 1: A key observation
If an event $X$ satisfies $X\subseteq\overline{X}$, then $X=\varnothing$.
Why: if some outcome $\omega$ were in $X$, then (because $X\subseteq\overline{X}$) it would also be in $\overline{X}$, i.e. not in $X$ — a contradiction. So $X$ has no outcomes.
Recall also: $\overline{\varnothing}=\Omega$ and $\overline{\Omega}=\varnothing$.
Step 2: Part $1^{\circ}$: $\;A+B=\overline{A}$
Finding the condition
Always $A\subseteq A+B$. Since $A+B=\overline{A}$, this gives $A\subseteq\overline{A}$, hence (Step 1) $A=\varnothing$.
Put $A=\varnothing$ back into the equality: the left side is $\varnothing+B=B$, the right side is $\overline{\varnothing}=\Omega$. So $B=\Omega$.
Check
If $A=\varnothing$ and $B=\Omega$: $A+B=\varnothing+\Omega=\Omega$ and $\overline{A}=\Omega$. ✓
\[ \boxed{A+B=\overline{A} \iff A=\varnothing \text{ and } B=\Omega.} \]
In words: $A$ must be impossible and $B$ must be certain.
Step 3: Part $2^{\circ}$: $\;AB=\overline{A}$
Finding the condition
Always $AB\subseteq A$. Since $AB=\overline{A}$, this gives $\overline{A}\subseteq A = \overline{\overline{A}}$. By Step 1 (with $X=\overline{A}$), $\overline{A}=\varnothing$, i.e. $A=\Omega$.
Put $A=\Omega$ back: the left side is $\Omega B = B$, the right side is $\overline{\Omega}=\varnothing$. So $B=\varnothing$.
Check
If $A=\Omega$ and $B=\varnothing$: $AB=\Omega\varnothing=\varnothing$ and $\overline{A}=\varnothing$. ✓
\[ \boxed{AB=\overline{A} \iff A=\Omega \text{ and } B=\varnothing.} \]
In words: $A$ must be certain and $B$ must be impossible.
Step 4: Part $3^{\circ}$: $\;A+B=AB$
If $A=B$, the equality holds
\[ A+B = A+A = A, \qquad AB = AA = A, \]
so $A+B=AB$ (idempotent laws).
If the equality holds, then $A=B$
Always $A\subseteq A+B$ and $A+B=AB\subseteq B$. Hence $A\subseteq B$.
In the same way $B\subseteq A+B=AB\subseteq A$, hence $B\subseteq A$.
Two inclusions in both directions mean $A=B$. $\blacksquare$
\[ \boxed{A+B=AB \iff A=B.} \]
In words: "at least one occurs" coincides with "both occur" only when the two events always occur together. If they could ever differ — one occurs and the other does not — then "at least one" would occur while "both" would not.
Step 5: Examples with a die
| Equality | Events | Left side | Right side | Holds? |
|---|---|---|---|---|
| $A+B=\overline{A}$ | $A=\varnothing,\ B=\Omega$ | $\Omega$ | $\Omega$ | ✓ |
| $A+B=\overline{A}$ | $A=\{1\},\ B=\{2,\dots,6\}$ | $\Omega$ | $\{2,\dots,6\}$ | ✗ |
| $AB=\overline{A}$ | $A=\Omega,\ B=\varnothing$ | $\varnothing$ | $\varnothing$ | ✓ |
| $A+B=AB$ | $A=B=\{2,4,6\}$ | $\{2,4,6\}$ | $\{2,4,6\}$ | ✓ |
| $A+B=AB$ | $A=\{2,4\},\ B=\{4,6\}$ | $\{2,4,6\}$ | $\{4\}$ | ✗ |
The second row shows that "$B$ is the complement of $A$" is not enough for $1^{\circ}$: $A$ itself must be impossible.
Result
| Equality | Possible exactly when | Meaning |
|---|---|---|
| $1^{\circ}\ A+B=\overline{A}$ | $A=\varnothing,\ B=\Omega$ | $A$ is impossible, $B$ is certain |
| $2^{\circ}\ AB=\overline{A}$ | $A=\Omega,\ B=\varnothing$ | $A$ is certain, $B$ is impossible |
| $3^{\circ}\ A+B=AB$ | $A=B$ | $A$ and $B$ are the same event |