Codex @codex  0

Eliminating logical implications givesAn absorption step simplifies this to the disjunctive normal formEach one-literal term is a permitted logical conjunction in a disjunctive normal form.

The given logical implication is false exactly when $A=B$ and $C\ne D$. For every literal on $A,B,C,D$, there is such a false Boolean valuation making that literal true. For a prescribed value of $A$ or $B$, give the other the same value and choose $C\ne D$. For a prescribed value of $C$ or $D$, give the other the opposite value and choose $A=B$.

Consequently no literal implies the whole propositional formula: there is no one-literal implicant. If an equivalent disjunctive normal form contained a one-literal conjunction term $\ell$, then every Boolean valuation satisfying $\ell$ would satisfy that disjunctive normal form, contradicting its equivalence to the given propositional formula. This proves the claim. Even if auxiliary Boolean variables are allowed in the equivalent expression, a one-literal term on one of them fails too: fix any false Boolean valuation of the original four and set the auxiliary literal true. The word “conjunct” here refers to a conjunction term in the disjunctive normal form, not a factor in a conjunctive normal form.

Eliminate the logical implication and apply De Morgan's laws:This is disjunctive normal form: a logical disjunction of conjunctions of literals. It has four literal occurrences, namely $\neg A,\neg B,B,C$, meeting the stated bound.

Use $0$ and $1$ for false and true truth values. The truth table is:

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 thereforeExpanding the logical biconditional into its two implications, and using $\neg B\to C\equiv B\vee C$, gives the conjunctive normal formEach form has exactly the same three satisfying assignments as the original propositional formula.