Confusion about there exist and forall Which of the following is a correct predicate logic statement for
"every natural number has [at least] one successor?"
\begin{align*} A: \quad & \forall x\exists y\left(\operatorname{succ}(x,y)\wedge\left(\exists z\operatorname{succ}(x,z)\implies\operatorname{equal}(y,z)\right)\right)\\
B: \quad &\forall x\exists y\left(\operatorname{succ}(x,y)\vee\left(\exists z\operatorname{succ}(x,z)\implies\operatorname{equal}(y,z)\right)\right)\\
C: \quad &\exists y\forall x\left(\operatorname{succ}(x,y)\wedge\left(\exists z\operatorname{succ}(x,z)\implies\operatorname{equal}(y,z)\right)\right)\\
D: \quad &\forall x\exists y\operatorname{succ}(x,y)
\end{align*}
I feel that both $A$ and $D$ are true. Is this right?
A: Statement $A$: for every natural number $x$, there exists a natural
number $y$ such that $y$ is the successor of $x$ and there exists
a natural number $z$ such that whenever $z$ is the successor of
$x$, $y$ and $z$ are equal (note that this last part is vacuously
true by taking $z=y$, so the whole statement becomes $\forall x\exists y\operatorname{succ}(x,y)$).
In that sense, $A= D$, but I would say that $D$ captures the
spirit of the question as intended.
