Let $\mathbb N$ denote the set of all non-negative integers, i.e. $0$, $1$, $2$, $3$, and so on.
Let $A$ be an arbitrary set and $R$ a subset of $A\times\mathbb N$. Furthermore, suppose that we have $$(\forall a\in A\colon (a, n)\in R)\implies (\forall b\in A\colon (b, n+1)\in R)$$ for all $n\in\mathbb N$. Does it follow that $$(c, m)\in R\implies (c, m+1)\in R$$ for all $c\in A$ and all natural numbers $m\in\mathbb N$?
No. Consider $A=\{1, 2\}$, and let $R=\{(1, 0)\}$.
Note that your first property has the form "$(\forall x . . .)\rightarrow (\forall x . . .)$", while your second property has the form "$\forall x(...\rightarrow ...)$". In general, "$\forall $" does not distribute across "$\rightarrow$".