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There are two languages $L_1=\{+\}$ with equation, where nonlogical symbol is binary function. There are formulas:

$$φ≡∃n∀x(n+x=x)∧∃n∀x(x+n=x)$$ $$ψ≡∃n∀x(n+x=x∧x+n=x)$$ There are theories $T_1={φ}$, $T_2={ψ}$ under language $L_1$ Could somebody please help me to find out if $T_2$ is [ordinary|conservative] extension of $T_1$?

Levitan
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  • $T_2$ has the same non-logical symbols as $T_1$, so why are you talking about conservative extensions? $\psi$ is strictly stronger than $\phi$ because an operation with a left-unit and a right-unit need not have a two-sided unit. So in a rather odd sense, the theory of $\psi$ is not conservative over the theory of $\phi$. – Rob Arthan Dec 18 '14 at 22:21

1 Answers1

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It is straightforward first-order logic to see that $\psi$ implies $\phi$ ("take the same $n$"), so that $T_2 \models T_1$.

Conversely, suppose that $M \models T_1$. Then there exist $n_l, n_r$ such that:

$$\forall x \in M: n_l + x = x = x + n_r$$

Now consider the expression $n_l + n_r$. According to the left equality, it equals $n_r$, while according to the right equality, it equals $n_l$. Therefore, $n_l = n_r$; that is to say:

$$\exists n \in M: \forall x \in M: n + x = x = x + n$$

which is precisely the condition for $M \models T_2$. Therefore, $T_1 \models T_2$ and the two theories are elementarily equivalent.

Lord_Farin
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