Let $\mathfrak A$ and $\mathfrak B$ are models of finite signature $\sigma$. Prove that $\mathfrak A$ and $\mathfrak B$ are isomorphic, if $\mathfrak A \equiv \mathfrak B$ and $\mathfrak A$ is finite model.
So what I made:
Let $\parallel\mathfrak A \parallel$ $\le$ $\parallel\mathfrak B\parallel$ and $\parallel \mathfrak A \parallel = m$. $ $ $\mathfrak A \equiv \mathfrak B$ so for all axioms $\varphi = (\exists y_1 ... \exists y_m)((y_1 \not = y_2)\wedge ... \wedge (y_{m-1} \not = y_m))$ performed $(\mathfrak A \models \varphi$ $\Leftrightarrow$ $\mathfrak B \models \varphi)$ $\Rightarrow$ $(\parallel\mathfrak A \parallel = \parallel\mathfrak B\parallel)$ $\Rightarrow \exists$ $f:A \rightarrow B$ is a bijection, where $\mathfrak A = <A, \sigma>$ and $\mathfrak B = <B, \sigma>$.
How to prove that $f$ is an isomorphism?
Could you verify my proof?
Let signature $\sigma = <f_1^{S_1},...,f_m^{S_m},P_1^{n_1},...,P_k^{n_k},c_1,...c_l>$ , where superscript is arity of function or relations. I wrote $\varphi = (\exists y_1...\exists y_q)((f_1(y_1,...,y_{S_1}) = y_j)\wedge...\wedge (f_m(y_1,...,y_{S_m}))\wedge(P_1^{n_1}y_1,...,y_{n_1}))\wedge...\wedge(P_k^{n_k}y_1,...,y_{n_k}))\wedge(c_1=y_{t_1})\wedge...\wedge(c_l = y_{t_l})\wedge(y_1 \not= y_2)\wedge...\wedge (y_{q-1} \not= y_q))$
After we signify this sentence in first model so that $\mathfrak A \models \varphi$($\gamma$ - is signification and $\gamma(y_i) = a_i$ , where $a_i \in A$). $\mathfrak A \equiv \mathfrak B \Rightarrow \mathfrak B \models \varphi$ when $\zeta$ ($\zeta$ is signification and $\zeta(y_i) = b_i$). Let $f:A \rightarrow B$ and $f(a_i) = b_i$.Then $f$ is our isomorphism.
Thanks!