Prove that the events 1^(°)…
Prove that the events \[ 1^{\circ}\ (A+B)(A+\overline{B})+(\overline{A}+B)(\overline{A}+\overline{B}) \qquad\text{and}\qquad 2^{\circ}\ (A+B)(\overline{A}+B)+(A+\overline{B})(\overline{A}+\overline{B}) \] are certain events.
Explanation of Method
An event is certain if it equals the whole sample space $\Omega$. Each product of two brackets here has a common term, so we use the rule $(X+Y)(X+Z)=X+YZ$ (the second distributive law). The leftover product is always of the form $B\overline{B}$ or $A\overline{A}$, which is impossible, so each product collapses to a single event. The two products in each sum turn out to be an event and its opposite, whose sum is $\Omega$.
Step 1: The key rule $(X+Y)(X+Z)=X+YZ$
Multiply out and absorb:
\[ (X+Y)(X+Z) = XX + XZ + YX + YZ = X + XZ + XY + YZ = X + YZ, \]
because $XX=X$, and $XZ\subseteq X$, $XY\subseteq X$ are absorbed by $X$. In words: "($X$ or $Y$) and ($X$ or $Z$)" means "$X$, or else both $Y$ and $Z$".
Step 2: Part $1^{\circ}$ — the first product
Common term $X=A$, with $Y=B$, $Z=\overline{B}$:
\[ (A+B)(A+\overline{B}) = A + B\overline{B} = A + \varnothing = A. \]
Step 3: Part $1^{\circ}$ — the second product
Common term $X=\overline{A}$:
\[ (\overline{A}+B)(\overline{A}+\overline{B}) = \overline{A} + B\overline{B} = \overline{A}. \]
Step 4: Part $1^{\circ}$ — the sum
\[ (A+B)(A+\overline{B})+(\overline{A}+B)(\overline{A}+\overline{B}) = A+\overline{A} = \Omega. \]
\[ \boxed{\text{Event } 1^{\circ} \text{ is certain.}} \]
In words: the first product means "$A$ occurs" (whatever $B$ does), the second means "$A$ does not occur". One of these always happens.
Step 5: Part $2^{\circ}$ — the two products
Now the common term is $B$ in the first product and $\overline{B}$ in the second:
\[ (A+B)(\overline{A}+B) = (B+A)(B+\overline{A}) = B + A\overline{A} = B, \]
\[ (A+\overline{B})(\overline{A}+\overline{B}) = (\overline{B}+A)(\overline{B}+\overline{A}) = \overline{B} + A\overline{A} = \overline{B}. \]
Step 6: Part $2^{\circ}$ — the sum
\[ (A+B)(\overline{A}+B)+(A+\overline{B})(\overline{A}+\overline{B}) = B+\overline{B} = \Omega. \]
\[ \boxed{\text{Event } 2^{\circ} \text{ is certain.}} \]
In words: "$B$ occurs" or "$B$ does not occur" — always true.
Step 7: A picture — which pieces does each bracket keep?
$A$ and $B$ split $\Omega$ into four pieces $AB$, $A\overline{B}$, $\overline{A}B$, $\overline{A}\,\overline{B}$. Each bracket removes exactly one piece:
| Bracket | Fails only when | Excludes the piece |
|---|---|---|
| $A+B$ | $A$ and $B$ both fail | $\overline{A}\,\overline{B}$ |
| $A+\overline{B}$ | $A$ fails and $B$ occurs | $\overline{A}B$ |
| $\overline{A}+B$ | $A$ occurs and $B$ fails | $A\overline{B}$ |
| $\overline{A}+\overline{B}$ | $A$ and $B$ both occur | $AB$ |
- $(A+B)(A+\overline{B})$ removes $\overline{A}\,\overline{B}$ and $\overline{A}B$, leaving $AB+A\overline{B}=A$.
- $(\overline{A}+B)(\overline{A}+\overline{B})$ removes $A\overline{B}$ and $AB$, leaving $\overline{A}$.
- Together they keep all four pieces — that is $\Omega$. Part $2^{\circ}$ works the same way with $B$ and $\overline{B}$.
Step 8: Truth-table check
| $A$ | $B$ | $(A+B)(A+\overline{B})$ | $(\overline{A}+B)(\overline{A}+\overline{B})$ | Event $1^{\circ}$ | $(A+B)(\overline{A}+B)$ | $(A+\overline{B})(\overline{A}+\overline{B})$ | Event $2^{\circ}$ |
|---|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 | 1 | 1 | 0 | 1 |
| 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 |
Both events equal $1$ in every row, i.e. they always occur. ✓
Result
| Event | Simplification | Result |
|---|---|---|
| $1^{\circ}$ | $(A+B)(A+\overline{B})=A,\ \ (\overline{A}+B)(\overline{A}+\overline{B})=\overline{A}$ | $A+\overline{A}=\Omega$ — certain |
| $2^{\circ}$ | $(A+B)(\overline{A}+B)=B,\ \ (A+\overline{B})(\overline{A}+\overline{B})=\overline{B}$ | $B+\overline{B}=\Omega$ — certain |