Show that overlineA+B C=overlineAC+overlineBC holds if and…

Show that $\overline{A+B}\,C=\overline{A}C+\overline{B}C$ holds if and only if $AC=BC$.

Explanation of Method

We simplify both sides with De Morgan's laws: the left side becomes "$C$ and neither $A$ nor $B$", the right side "$C$ and not both $A$ and $B$". Then we split $C$ into four non-overlapping pieces according to $A$ and $B$. The left side is one piece, the right side is three pieces, so they are equal exactly when the two extra pieces are empty — and that is precisely the condition $AC=BC$.

Step 1: Simplify the left side

By De Morgan, $\overline{A+B}=\overline{A}\,\overline{B}$, so

\[ \overline{A+B}\,C = \overline{A}\,\overline{B}\,C \quad(\text{$C$ occurs, and neither $A$ nor $B$}). \]

Step 2: Simplify the right side

Take $C$ outside the bracket and use De Morgan $\overline{A}+\overline{B}=\overline{AB}$:

\[ \overline{A}C+\overline{B}C = (\overline{A}+\overline{B})C = \overline{AB}\,C \quad(\text{$C$ occurs, and not both $A$ and $B$}). \]

Step 3: Split $C$ into four pieces

Every outcome of $C$ lies in exactly one of

Piece	In the left side $\overline{A}\,\overline{B}C$?	In the right side $\overline{AB}\,C$?
$ABC$	no	no
$A\overline{B}C$	no	yes
$\overline{A}BC$	no	yes
$\overline{A}\,\overline{B}C$	yes	yes

$\Omega$ $A$ $B$ $\overline{A}\,\overline{B}C$ $A\overline{B}C$ $\overline{A}BC$

left side $\overline{A+B}\,C=\overline{A}\,\overline{B}C$
the extra pieces $A\overline{B}C$, $\overline{A}BC$ in the right side

Inside $C$: the left side is only the green piece; the right side is green + orange. They are equal exactly when both orange pieces are empty.

The two sides differ only in the pieces $A\overline{B}C$ and $\overline{A}BC$. Therefore

\[ \overline{A+B}\,C=\overline{A}C+\overline{B}C \iff A\overline{B}C=\varnothing \ \text{ and } \ \overline{A}BC=\varnothing. \tag{$*$} \]

Step 4: Condition $(*)$ is the same as $AC=BC$

Split $AC$ and $BC$ into pieces as well:

\[ AC = ABC + A\overline{B}C, \qquad BC = ABC + \overline{A}BC. \]

If $(*)$ holds, both extra pieces vanish, so $AC = ABC = BC$.
If $AC=BC$, multiply by $\overline{B}$: $A\overline{B}C = BC\overline{B} = \varnothing$. Multiply by $\overline{A}$: $\overline{A}BC = AC\overline{A} = \varnothing$. So $(*)$ holds.

Combining with Step 3:

\[ \boxed{\overline{A+B}\,C=\overline{A}C+\overline{B}C \iff AC=BC.} \quad\blacksquare \]

Step 5: In words

$AC=BC$ means: when $C$ occurs, $A$ occurs exactly when $B$ occurs. Then, inside $C$, "neither $A$ nor $B$" and "not both $A$ and $B$" say the same thing, because the mixed cases (exactly one of $A$, $B$) never happen.

Step 6: Example with a die

$\Omega=\{1,\dots,6\}$, $C=\{1,2,3\}$.

$A$	$B$	$AC$	$BC$	Left side	Right side
$\{1,4\}$	$\{1,5,6\}$	$\{1\}$	$\{1\}$	$\{2,3\}$	$\{2,3\}$ ✓
$\{1,2\}$	$\{1\}$	$\{1,2\}$	$\{1\}$	$\{3\}$	$\{2,3\}$ ✗

In the first row $AC=BC$ and the equality holds; in the second row $AC\ne BC$ (outcome $2$ is in $A\overline{B}C$) and the equality fails.

Result


Left side	$\overline{A+B}\,C=\overline{A}\,\overline{B}C$
Right side	$\overline{A}C+\overline{B}C=\overline{AB}\,C$
Equal exactly when	$A\overline{B}C=\overline{A}BC=\varnothing$, i.e. $AC=BC$