Further to MJD's, Dario's and Cheerful Parsnip's comments and answers, the sentence
- Every $A$ has a unique $B$ (❔)
is arguably ambiguous between these clearer constructions:
Every $\boldsymbol A$ has exactly one $\boldsymbol B$
$\boldsymbol B$ is uniquely determined by $\boldsymbol A$
(This reading corresponds to the common technical usage of ‘unique’ to mean sole.
Notice that with this reading:
- $\text“x-3=0$ has a unique root, $3\text”$ implies neither “The number $3$ is a unique root” nor “The root $3$ is unique to the equation $x-3=0\text”$ !
- $\text“-2$ and $2$ are unique roots of $x^2=4\text”$ doesn't imply $\text“x^2=4$ has a unique root” !)
Every $\boldsymbol A$ has a $\boldsymbol B$ that has no duplicate
(In a technical setting, this reading is a bit liberal, and corresponds to the everyday meaning of ‘unique’ as one of a kind.)
To wit, consider the sentence “Every bin has a unique score” (bins are separated by indentation and scores by commas).
\begin{gather}
7 &7 &8 &9\tag{1}\\
6,0 &7,0 &8,0 &9,0\tag{2}
\end{gather}
On the other hand, the conjunction of Readings 1 and 2 is logically equivalent to this sentence:
Every $\boldsymbol A$ has a distinct $\boldsymbol B$
Each $\boldsymbol A$ has exactly one $\boldsymbol {B,}$ which has no duplicate
\begin{gather}
6&&7&&8&&9\tag{1,2,3}\\
\end{gather}
It turns out that the Question's four statements have four distinct meanings!
$(S_1)\quad$ “Every boy owns a unique shirt.”
$(S_2)\quad$ “Every shirt belongs to a unique boy.”
$(S_3)\quad$ “No two boys own the same shirt.”
Every shirt belongs to at most one boy. $$∀s,b_1,b_2\;\Big(P(s,b_1)∧P(s,b_2)\implies b_1=b_2\Big).$$
$(S_4)\quad$ “No two shirts belong to the same boy.”
Every boy owns at most one shirt. $$∀b,s_1,s_2\;\Big(P(s_1,b)∧P(s_2,b)\implies s_1=s_2\Big).$$
$(S_1)$ and $(S_2),$ based on Meaning 1 (alternatively: Meaning 2) in the previous section, are clearly inequivalent;
since $(S_3)$ and $(S_4)$ each allows some boy to own no shirt, neither is equivalent to $(S_1);$
since $(S_3)$ and $(S_4)$ each allows some shirt to have no owner, neither is equivalent to $(S_2);$
$(S_3)$ and $(S_4)$ are clearly inequivalent.