Let $\varphi(x_1,\dots,x_n)$ be a $\Delta_0$-formula, i.e. a formula in which every quantifier is bounded. I want to prove that $$ \text{PA}\vdash\varphi(\overline{n_1},\dots,\overline{n_k}) \iff \mathcal{N}\models\varphi[n_1,\dots,n_k] $$ holds for all natural numbers $n_1,\dots,n_k\in\mathbb{N}$, where $\mathcal{N}$ denotes the standard model and where $\overline{n_i}$ denotes the numeral of $n_i$.
I can see how the above equivalence holds for all atomic formulas, but I can't see how it holds for formulas of the form $\neg\varphi$, where $\varphi$ is assumed to already satisfy the above equivalence. If $\text{PA}\vdash\neg\varphi$, then surely $\mathcal{N}\models\neg\varphi$, but for the converse: if $\mathcal{N}\models\neg\varphi$, I can only conclude that $\text{PA}\not\vdash\varphi$, which, as I see it, does not necessarily imply that $\text{PA}\vdash\neg\varphi$.
Any help would be greatly appreciated.
The key point is that an atomic formula of PA says that one specific natural number is equal to another specific natural number, or that one specific natural number is less than another specific natural number. For example, the formula could look like $3 = 8$ or $2 \leq 10$. If such a formula is false, PA contains sufficient axioms to prove it is false.
For example, PA proves that: $3 =8$ if and only if $2 = 7$, if and only if $1 = 6$, if and only if $0 =5$, and because PA proves $0 \not = 5$, PA proves $3 \not = 8$. There is no induction here: because $3$ and $8$ are specific numbers, we can include the entire chain of equivalences in a single proof.
It is not hard to go from atomic formulas and negated atomic formulas to general quantifier free formulas. Bounded quantifiers are handled by replacing them with appropriate conjunctions or disjunctions, which can be done because the bound on the quantifer is again a specific number in the case at hand.