I'm trying to find example of implementation $M$ of language $L$ such that $M \models \varphi_1 \land \varphi_2 \land \varphi_3 \land \varphi_4$
Where $L = \{•, \blacksquare, n\}$ is language with equality symbol, where all 3 symbols are functional with airities 2, 2, 0 and formulas:
$\varphi_1 \equiv ∀a∀b∀c(a • b = b • a ∧ (a • b) • c = a • (b • c) ∧ a • n = a ∧ a \blacksquare b = b \blacksquare a)$, $\varphi_2 \equiv ∀a∀b(∃c(c • (a \blacksquare b) = a) ∧ ∃c(c • (a \blacksquare b) \ne n))$, $\varphi_3 ≡ ∀a∀b∀c∀d∀e(((c • d) • (a \blacksquare b) = a ∧ (c • e) • (a \blacksquare b) = b) → c = n)$, $\varphi_4 ≡ ∃a∃b(a \blacksquare b \ne a ∧ a \blacksquare b \ne b)$.
but still with no success. Could anybody suggest me some flow of proof?
Update: Maybe somebody knows how to do this for $M \models \varphi_1 \land \varphi_2$
Consider $M=\Bbb{Q}$, $•^M=+$, $\blacksquare^M=\times$ and $n^M=0$. You can check that $\varphi_1$, $\varphi_2$ and $\varphi_4$ are true in $M$. Also, $\varphi_3$ is vacuously true since there is no rational $c$ such that $c+d+ab=a$ and $c+e+ab=b$ for every $a,b,d,e\in\Bbb Q$.