A conservative extension There are two languages $L_1 = \{+\}$, $L_2 = \{\cdot\}$, $L_3 = \{+, \cdot\}$ with equation, where both nonlogical symbols are binary functions.
There are formulas:
$$\varphi \equiv \exists n \forall x (n+x = x) \wedge \exists n \forall x(x + n = x) $$
$$\psi \equiv \exists n \forall x (n+x=x \wedge x+n=x) $$
$$\xi \equiv \forall x \forall y \forall z(x \cdot (y+z) = (x\cdot y) + (x\cdot z)) $$
$$\zeta \equiv \forall x \forall y \forall z(x + y = y + x \wedge (x+y)+z = x+(y+z)) $$
$$\eta \equiv \forall x \exists y \forall z((x+y)+z = z)$$
There are theories $T_1 = \{\varphi\}$, $T_2 = \{\psi\}$ under language $L_1$, theory $T_3 = \varnothing$ under lanugage $L_2$, $T_4=\{\psi, \xi, \zeta, \eta\}$, $T_5=\{\xi\}$ under language $L_3$
Could somebody please help me to find out if $T_4$ or $T_5$ is a conservative extension of $T_3$?
Update:
I stucked with finding if $T_2$ is [conservative|ordinary] extension of $T_1$. Seems like mathlogics is not my field. Could somebody help me here?
 A: $T_4$ is not conservative over $T_3$ but $T_5$ is. For the first claim, consider the following statement in $L_2$:


*

*$\chi \equiv \forall x, y(x \cdot y = x)$


Suppose $\chi$ were true. From $\xi$ it would follow that $\forall x(x = x + x)$. Now, consider an arbitrary $x$. By $\psi$ there is some $n$ such that $x + n = x$. By $\eta$ there is some $y$ such that $(x + y) + n = n$, and thus $(x + y) = n$ by $\psi$. So $(x + y) + x = n +x = x$ by $\psi$. But $(x+ y) +x = (x+ x) + y$ by $\zeta$. So $x = (x+ y) + x = (x+ x) + y = x + y = n$ (the second to last equality holds because $\forall x(x+ x = x)$). Since $x$ was arbitrary, it follows that $\exists n\forall x(x = n)$. This establishes:


*

*$T_4\vdash \chi \to \exists n\forall x(x = n)$


Since "$\chi \to \exists n\forall x(x = n)$" is a statement in $L_2$ and not a logical truth, $T_4$ is not conservative over $T_3$.
For the second claim, note that given any model $\mathcal M$ of $L_2$ we can expand it to a model $\mathcal M'$ of $L_3$ by letting $x + y = x$. Then it is clear that $\mathcal M'\vDash \xi$. 
On the update: Let $m$ be such that $m + x = x$ and $n$ such that $x + n = x$. Then $n = m + n = m$. So $\psi$ follows from $\phi$. 
