Comparing semiring of formulas and Lindenbaum algebra

Question

This is motivationally related to an earlier question of mine.

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Given a first-order theory $T$, let $\widehat{D}(T)$ be the semiring defined as follows:

• Elements of $\widehat{D}(T)$ are equivalence classes of formal $\mathbb{N}$-linear combinations of first-order formulas in the language of $T$.

• The equivalence relation is the finest one satisfying the following properties (writing "$[\varphi]$" for the class of $\varphi$):

  • $[\varphi\vee\psi]+[\varphi\wedge\psi]=[\varphi]+[\psi]$. (The slightly odd phrasing is needed since we don't have $-1$ as a scalar.)

  • If $\varphi,\psi$ share no variables in common then $[\varphi]\cdot [\psi]=[\varphi\wedge\psi]$.

  • $[\varphi]=n[\psi]$ whenever there is a formula $\theta$ which $T$ proves defines a $n$-to-one surjection from the set of tuples satisfying $\varphi$ to the set of tuples satisfying $\psi$.

Note that for sentences, "$T$-provable" and "$T$-disprovable" imply "$=1_{\widehat{D}(T)}$" and $=0_{\widehat{D}(T)}$" respectively. More interestingly, taking $n=1$ in the third bulletpoint we get that if $\varphi$ and $\psi$ are sentences (so the corresponding sets of tuples are either $\emptyset$ or $\{\emptyset\}$) then $[\varphi]=[\psi]$ whenever $T\vdash\varphi\leftrightarrow\psi$ (via $\theta\equiv\top$). However, the converse isn't clear to me.

Suppose $\varphi,\psi$ are sentences in the language of $T$. If $[\varphi]=[\psi]$, must we have $T\vdash\varphi\leftrightarrow\psi$?

I generally want to understand how $\widehat{D}(T)$ can (if at all!) disagree with the Lindenbaum algebra of $T$. For example, let $\varphi\triangleleft\psi$ iff there are formulas $\chi_1,...,\chi_n$ such that $[\varphi]+[\chi_1]+...+[\chi_n]=[\psi]$. Then do we have $T\vdash\varphi\rightarrow\psi$ iff $[\varphi]\triangleleft[\psi]$? For now, however, I think the question above is the right thing to focus on.

Answer 1

Yes, if $[\varphi]=[\psi]$ for sentences $\varphi$ and $\psi$, then $T\vdash\varphi\mathrel{\leftrightarrow}\psi$.

Indeed, for this we assume that $[\varphi]=[\psi]$ and fix a model $\mathfrak{M}\models T$ to show that $\mathfrak{M}\models \varphi\iff\mathfrak{M}\models \psi$.

Notice that the latter is equivalent to the assertion about the equality of the cardinalities of the sets of tuples satisfying $\varphi$ and $\psi$ in $\mathfrak{M}$. In order to prove that it is sufficient to observe that the equivalence relation on $T$-formulas $$\theta \sim_{\mathfrak{M},\text{crd}} \eta\iff |\{\vec{x}\mid \mathfrak{M}\models \theta(\vec{x})\}|=|\{\vec{y}\mid \mathfrak{M}\models \eta(\vec{y})\}|$$ satisfies the three bullet points from your question (and hence showing that $[\theta]=[\eta]$ implies $[\theta]_{\mathfrak{M},\text{crd}}=[\eta]_{\mathfrak{M},\text{crd}}$).