# SHOW that there are infinitely many equivalence classes of formulas

Let $\mathcal{Q}$ denote the additive group of rational numbers, i.e. the structure $\left<\mathbb{Q}; +; 0\right>$. Let $\mathcal{L}$ be the language of $\mathcal{Q}$ and let $T$ be the complete theory of $\mathcal{Q}$. By considering automorphisms of $\mathcal{Q}$ given that every formula in $F_1(\mathcal{L})$ is $E_1(T)$-equivalent to exactly one of the four formulas

1. $v_1 = v_1$
2. $v_1 = 0$
3. $-v_1 = 0$
4. $-v_1 = v_1$

prove that there are infinitely many $E_2(T)$ - equivalence classes of formulas in $F_2(T)$?

Fn(L) denotes the set of all L-formulas ϕ with FrVar(ϕ)⊆{v1,...,vn}


En(T) denotes the binary relation on Fn(L) defined by

(ψ ,ϕ)∈En(T)⟺T⊨∀v1,...,vn(ϕ(v1,...,vn)⟺ψ(v1,...,vn))

I have proved that considering automorphisms of $\mathcal{Q}$, every formula is equivalent to exactly one of the above formulas, but cannot prove the last part - infinitely many equivalence classes. Please help.

• I feel like the title of this question could be better-worded; I know quite a lot of group theory, but nothing of model theory or wherever this question comes from. – Mr. Chip Mar 30 '13 at 14:18
• Thanks for improving the formatting of this question. This is from Model theory - covering concepts of Isomorphism, Non-Example, Similarity Type, Structure, Class of all Structures, Embedding, Sub-structure, Compactness theorem, etc. – user70161 Mar 30 '13 at 14:26
• Can you share us what are $E_1(T),\ E_2(T),\ F_1(\mathcal L)$ and $F_2(T)$? – Berci Mar 30 '13 at 15:27
• Fn(L) denotes the set of all L-formulas ϕ with FrVar(ϕ)⊆{v1,...,vn} En(T) denotes the binary relation on Fn(L) defined by (ψ ,ϕ)∈En(T)⟺T⊨∀v1,...,vn(ϕ(v1,...,vn)⟺ψ(v1,...,vn)) – user70161 Mar 31 '13 at 0:25
• Consider the formulas "$n\cdot v_1=v_2$", where "$n\cdot v_1$" is formally "$v_1+\ldots+v_1$" ($n$ times). I think this works but it's late here so I might be wrong. – Apostolos Mar 31 '13 at 0:58

HINT: For each $n\in\omega$ consider the formulas $$\varphi_n\equiv(\underbrace{v_1+\ldots+v_1}_n=v_2).$$ These formulas are elements of $F_2(\mathcal{L})$ and for $n\neq m$ you can show that $(\varphi_n,\varphi_m)\notin E_2(T)$. Hence the elements of $\{[\varphi_n] : n\in\omega\}$ are distinct ($[\varphi_n]$ being the corresponding $E_2(T)$-equivalence class).
• @AmarChaman: Really? This is quite easy: Take $n\neq m$, take $\frac{1}{n}$ as $v_1$ and $1$ as $v_2$. Then $\mathcal{Q}\models \underbrace{\frac{1}{n}+\ldots+\frac{1}{n}}_n=1$. On the other hand $\mathcal{Q}\models \underbrace{\frac{1}{n}+\ldots+\frac{1}{n}}_m\neq 1$. Hence $\mathcal{Q}\models\exists v_1,v_2(\underbrace{v_1+\ldots+v_1}_n=v_2\land\lnot(\underbrace{ v_1+\ldots+v_1 }_m =v_2))$. – Apostolos Apr 1 '13 at 12:07