Is there a mistake in this Model Theory example? This is taken from towards the end of Example 14.36 in Open Logic Project on p. 203 (Release 2020-06-25).
We have a first-order language $\mathcal{L} = \{ a, b, f, R \}$, where $a$ and $b$ are constant symbols, $f$ is a two-place function symbol and $R$ is a two-place predicate symbol. Its model is $\mathfrak{M}$, defined by:
\begin{align*}
&|\mathfrak{M}| = \{1,2,3,4\} \\
&a^{\mathfrak{M}} = 1 \\
&b^{\mathfrak{M}} = 2 \\
&f^{\mathfrak{M}}(x,y) = x + y \text{ if $x + y \leq 3$ and $= 3$ otherwise.} \\
&R^{\mathfrak{M}} = \{(1,1), (1,2), (2,3), (2,4)\},
\end{align*}
together with a variable assignment $s(v) = 1$ for every variable. The $x$-variants of $s$ are $s_1'(x) = 1$, $s_2'(x) = 2$, $s_3'(x) = 3$ and $s_4'(x) = 4$.
Now the example claims that
$$
\mathfrak{M}, s \nvDash \exists x(R(a,x) \wedge\forall y R(x,y)).
$$
The reasoning is:

Just to set things straight, the way I understand the definition of variants of variable assignments makes $s_1'(y) = 4$ and $1$ for all otehr variables and $s_2'(y) = 4$, $2$ if the variable is $x$ and $1$ for all other variables, since any $y$-variant of a variable assignment can differ from the "original" variable assignment at most in what it assigns to $y$.
Now the claim that $\mathfrak{M}, s_1' \nvDash R(x,y)$ is true, since $(1,4) \notin R^{\mathfrak{M}}$, but $\mathfrak{M}, s_2' \vDash R(x,y)$, since $(2,4)$ is an element of $R^{\mathfrak{M}}$! Therefore, the example contains a mistake. I can't say that the conclusion $\mathfrak{M}, s \nvDash \exists x(R(a,x)) \wedge\forall y R(x,y))$ is wrong though, since I haven't checked all $y$-variants of $s_2$. You may do that in comments, if you please. :)
Am I correct or not?
 A: With an abuse of notation, the claim ask for the truth-value of sentence $∃x(R(1,x)) ∧ ∀yR(x,y))$ in model $\mathfrak M$.
We have that $(1,1),(1,2) ∈ R$, and thus, the only candidates for $x$ are $1$ and $2$.
This is the meaning of the author's claim:

The only $x$-variant $s'_i$ of $s$ such that $\mathfrak M, s'_i \vDash R(a,x)$ are $s'_1$ and $s'_2$,

exactly because $s'_1(x)=1$ and $s'_2(x)=2$.
For the next step, we have that the formula to be satisfied by $s$ is universally quantified: $\forall x R(x,y)$. We have found a couple of $x$-variants of $s$ that satisfy the left conjunct: $s'_1$ and $s'_2$.
Now we have to check that at least one of them satisfies $\forall x R(x,y)$ and to do this we have to consider every $y$-variant of each of them.
To have $\mathfrak M, s'_1 \vDash ∀yR(x,y)$ we must have that, for every $y$-variant $s''_1$ of $s'_1$ we must have $\mathfrak M, s''_1 \vDash R(x,y)$.
But consider the $y$-ariant $s^*$ such that $s^*(x)=1$ and $s^*(y)=3$. We have:


$\mathfrak M, s^* \nvDash R(x,y)$,


because $(1,3) \notin R^{\mathfrak M}$. An the same for $(1,4)$.
The same reasoning applies to $s'_2$: $(2,1), (2,2) \notin R^{\mathfrak M}$.

So, yes, there is a mistake in the author's explanantion: the $y$-variant using $(1,4)$ will work, because $(1,4) \notin R^{\mathfrak M}$, but that using $(2,4)$ will not do, because $(2,4) \in R^{\mathfrak M}$.

