= Solution
Use $0$ and $1$ for false and true <truth values>. The <truth table> is:
|| $A$
|| $B$
|| $C$
|| $A\leftrightarrow B$
|| $\neg B\to C$
|| $\varphi$
| $0$
| $0$
| $0$
| $1$
| $0$
| $0$
| $0$
| $0$
| $1$
| $1$
| $1$
| $1$
| $0$
| $1$
| $0$
| $0$
| $1$
| $0$
| $0$
| $1$
| $1$
| $0$
| $1$
| $0$
| $1$
| $0$
| $0$
| $0$
| $0$
| $0$
| $1$
| $0$
| $1$
| $0$
| $1$
| $0$
| $1$
| $1$
| $0$
| $1$
| $1$
| $1$
| $1$
| $1$
| $1$
| $1$
| $1$
| $1$
The two satisfying rows with $A=B=1$ combine by eliminating $C$, while the remaining satisfying row has $A=B=0,C=1$. An equivalent <disjunctive normal form> is therefore
$$
\boxed{(A\land B)\lor(\neg A\land\neg B\land C).}
$$
Expanding the <logical biconditional> into its two implications, and using $\neg B\to C\equiv B\lor C$, gives the <conjunctive normal form>
$$
\boxed{(\neg A\lor B)\land(A\lor\neg B)\land(B\lor C).}
$$
Each form has exactly the same three <satisfying assignments> as the original <propositional formula>.
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