The statement “Everyone has exactly one mother” requires that for every individual $x$, there exists a unique $y$ such that $\texttt{mother}(y,x)$ holds. This entails:
- Existence: $\forall x \, \exists y \, \texttt{mother}(y,x)$
- Uniqueness: $\forall x \, \forall y \, \forall z \, \bigl( \texttt{mother}(y,x) \land \texttt{mother}(z,x) \rightarrow y = z \bigr)$
Option A
$$
\forall x \, \exists y \, \exists z \, \bigl( \texttt{mother}(y,x) \land \lnot \texttt{mother}(z,x) \bigr)
$$
This formula is satisfied if, for every $x$, there is some mother and some non-mother. It does not prohibit multiple mothers.
To demonstrate its inadequacy, consider the following countermodel:
$$
\begin{array}{l}
\text{Domain: } \{p, m_1, m_2, r\} \\
\text{Motherhood facts: } \\
\quad \texttt{mother}(m_1, p) \\
\quad \texttt{mother}(m_2, p) \\
\quad \lnot \texttt{mother}(r, p)
\end{array}
$$
Here, $p$ has two distinct mothers, violating “exactly one”. However, for $x = p$, choose $y = m_1$ and $z = r$: $\texttt{mother}(m_1, p) \land \lnot \texttt{mother}(r, p)$ is true. Thus, the formula holds, even though the intended meaning is false. Hence, Option A is incorrect.
Option B
$$
\forall x \, \exists y \, \Bigl[ \texttt{mother}(y,x) \land \forall z \, \bigl( \texttt{noteq}(z,y) \rightarrow \lnot \texttt{mother}(z,x) \bigr) \Bigr]
$$
This asserts: for each $x$, there is a $y$ such that:
- $y$ is a mother of $x$, and
- every $z \ne y$ is not a mother of $x$.
This precisely captures existence and uniqueness.
Suppose, for contradiction, that some $p$ has two mothers $m_1 \ne m_2$. Then for $x = p$, no choice of $y$ can satisfy the condition:
- If $y = m_1$, then for $z = m_2$, $\texttt{noteq}(m_2, m_1)$ is true but $\texttt{mother}(m_2, p)$ is also true, violating the implication.
- Similarly for $y = m_2$.
Thus, the formula is false in any model with multiple mothers, and true only when each person has exactly one mother.Therefore, Option B is correct.
Option C
$$
\forall x \, \forall y \, \Bigl[ \texttt{mother}(y,x) \rightarrow \exists z \, \bigl( \texttt{mother}(z,x) \land \lnot \texttt{noteq}(z,y) \bigr) \Bigr]
$$
Note that $\lnot \texttt{noteq}(z,y)$ is equivalent to $z = y$. Hence, the consequent becomes:
$$
\exists z \, \bigl( \texttt{mother}(z,x) \land z = y \bigr) \equiv \texttt{mother}(y,x)
$$
Thus, the entire formula simplifies to:
$$
\forall x \, \forall y \, \bigl( \texttt{mother}(y,x) \rightarrow \texttt{mother}(y,x) \bigr)
$$
which is a tautology.
Counterexample 1 (no mother):
- Let the domain be $\{p\}$, and assume $\lnot \texttt{mother}(y,p)$ for all $y$. The implication is vacuously true for all $y$, so the formula holds — yet $p$ has no mother.
Counterexample 2 (two mothers):
- Let $\texttt{mother}(m_1,p)$ and $\texttt{mother}(m_2,p)$ with $m_1 \ne m_2$. For $y = m_1$, the consequent is satisfied by $z = m_1$; similarly for $y = m_2$. The formula holds despite the violation of uniqueness.
Hence, Option C is incorrect.
Option D
$$
\forall x \, \exists y \, \Bigl[ \texttt{mother}(y,x) \land \lnot \exists z \, \bigl( \texttt{noteq}(z,y) \land \texttt{mother}(z,x) \bigr) \Bigr]
$$
This states: for each $x$, there is a $y$ such that:
- $y$ is a mother of $x$, and
- there is no $z \ne y$ that is also a mother of $x$.
Using standard quantifier negation:
$$
\lnot \exists z \, \bigl( \texttt{noteq}(z,y) \land \texttt{mother}(z,x) \bigr)
\equiv
\forall z \, \bigl( \texttt{noteq}(z,y) \rightarrow \lnot \texttt{mother}(z,x) \bigr)
$$
Thus, Option D is logically equivalent to Option B.
In the countermodel where $p$ has two mothers $m_1 \ne m_2$, any candidate $y$ (say $m_1$) fails because $z = m_2$ satisfies $\texttt{noteq}(z,y) \land \texttt{mother}(z,x)$. Hence, the inner negation is false, and the formula does not hold as required.Therefore, Option D is correct.
$$
\color{lime} \boxed{\text{Options B and D are correct}}
$$
