Is that possible to define an explict structure $\mathfrak{A}$ such that $\mathfrak{A}$ is a substructure of $\mathfrak{B}$ which $\mathfrak{A}$ is isomorphic to the natural numbers $\mathfrak{N} = (\mathbb{N},0,S,+,\cdot,E,<)$.
Where $\mathfrak{B} \vDash N$ where $N = \{ N_1, ..., N_{11} \}$.
\begin{align*} &~\hspace{10mm} \mathbf{The~Axioms~of~}N \\ &N_1.~(\forall x) \neg S x = 0. \\ &N_2.~(\forall x)(\forall y) [S x = S y \rightarrow x = y]. \\ &N_3.~(\forall x) x + 0 = x. \\ &N_4.~(\forall x)(\forall y) x + S y = S(x + y). \\ &N_5.~(\forall x) x \cdot 0 = 0. \\ &N_6.~(\forall x)(\forall y) x \cdot S y = (x \cdot y) + x. \\ &N_7.~(\forall x) x E 0 = S 0. \\ &N_8.~(\forall x)(\forall y) x E (S y) = (x E y) \cdot x. \\ &N_9.~(\forall x) \neg x < 0. \\ &N_{10}.~(\forall x)(\forall y) [x < S y \leftrightarrow (x < y \vee x = y)]. \\ &N_{11}.~(\forall x)(\forall y) [(x < y) \vee (x = y) \vee (y < x)]. \end{align*}
Well, I am quite stuck in the beginning. It is even hard for me to imagine what will $\mathfrak{B}$ would be like and I know that $\mathfrak{B}$ cannot be exactly same as $\mathfrak{N}$ since, $\mathfrak{N}$ does not have a proper substructure. So, is that possible to define an explict structure $\mathfrak{A}$ such that $\mathfrak{A}$ is a substructure of $\mathfrak{B}$ which $\mathfrak{A}$ is isomorphic to the natural numbers $\mathfrak{N} = (\mathbb{N},0,S,+,\cdot,E,<)$.