If $\mathcal{L} = \{P, R, f, g, c_0, c_1\}$, where $P$ is a unary predicate, $R$ is a binary predicate, and $f$ and $g$ are binary function symbols. Let $\mathcal{M} = (D, \mathcal{I})$ be an $\mathcal{L}$-structure with $D = \mathbb{R}$, such that $\mathcal{I}(P) = \mathbb{N}$, $\mathcal{I}(R)$ is the usual larger-than (>) relation on $\mathbb{R}$, $\mathcal{I}(f)$ is the usual addition on $\mathbb{R}$, $\mathcal{I}(g)$ is the usual multiplication on $\mathbb{R}$, $\mathcal{I}(c_0) = 0$, and $\mathcal{I}(c_1) = 1$. Finally, let $s$ be a variable assignment over $\mathcal{M}$ with $s(x) = 5$ and $s(y) = 3$ (where $x$ and $y$ are distinct variables). What is the semantic value of this following $\mathcal{L}$-formula (using $Val_{\mathcal{M},s}$):
$∀x∀y \, \big(R(x, c_0) \to ∃z \, (P(z) \land R(g(z, x), y)) \big)$
I know that the first line of the solution says $Val_{\mathcal{M},s}\big(∀x∀y \, (R(x, c_0) \to ∃z \, (P(z) \land R(g(z, x), y)))\big) = \mathbf{1}$. I don't understand why you assume it is false before evaluating the $\mathcal{L}$-formula? (i.e. why is $\, = \mathbf{1}$ and not $ \, = \mathbf{0}$).
The interpretation of $\forall x \forall y \, \big(R(x,c_0) \to \exists z \, (P(z) \land R(g(z,x),y))\big)$ in $\mathcal{M}$ is $\mathsf{true}$ (and not $\mathsf{false}$), i.e. $\mathit{Val}_\mathcal{M} \big(\forall x \forall y \, (R(x,c_0) \to \exists z \, (P(z) \land R(g(z,x),y)))\big) = \mathsf{true}$, because for every variable assignment $s$ one has $\mathit{Val}_{\mathcal{M},s} \big(R(x,c_0) \to \exists z \, (P(z) \land R(g(z,x),y))\big) = \mathsf{true}$. Indeed, for whatever value of $x$ and $y$ in $\mathbb{R}$, the interpretation in $\mathcal{M}$ of the open formula $R(x,c_0) \to \exists z \, (P(z) \land R(g(z,x),y))$ is: "if $x > 0$ then there exists a natural number $z$ such that $z × x > y$", which is true because it just states the Archimedean property of $\mathbb{R}$.