= Solution
Assume first that $\phi$ is false on the all-true and all-false <Boolean valuations>. Use the truth-table normal form from part (ii). Every satisfying <Boolean valuation> on the finitely many <propositional variables> must have at least one true <propositional variable> and at least one false <propositional variable>. Its <minterm> therefore has the form
$$
C_s\wedge\neg D_s,
$$
where $C_s$ conjoins its true <propositional variables> and $D_s$ disjoins its false <propositional variables>; both lists are nonempty. This is equivalent to $\neg(C_s\to D_s)$. If the satisfying <Boolean valuations> are indexed by $1,\ldots,N$, <De Morgan's laws> gives
$$
\boxed{\phi\equiv\bigvee_{i=1}^N\neg(C_i\to D_i)
\equiv\neg\bigwedge_{i=1}^N(C_i\to D_i).}
$$
If there are no satisfying <Boolean valuations>, take $N=1$ and $C_1=D_1=p$: the result is $\neg(p\to p)$, again always false.
Conversely, on the all-true <Boolean valuation> every nonempty $D_i$ is true, so every $C_i\to D_i$ is true. On the all-false <Boolean valuation> every nonempty $C_i$ is false, so every <logical implication> is again true. Negating their <logical conjunction> gives false on both <Boolean valuations>. <Logical equivalence> transfers these values to $\phi$. The usual nonempty meaning of <logical conjunction> and <logical disjunction> of <propositional variables> is used here; allowing empty $C_i$ or $D_i$ without qualification would destroy this converse.
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