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Let us forget about function symbols for simplicity. $\def\MMM{\mathcal{M}} \def\LLL{\mathcal{L}} \def\DDD{\mathcal{D}}$

I've been wondering whether the usual definition of a model in first-order logic is equivalent to the following: The usual definition of a model is as follows:

Definition: A model $\MMM$ for a language $\LLL$ is tuple $(\DDD, \mu)$ where

  1. $\DDD$ is a superset of the constants in the language.
  2. For any predicate $P$ of arity $n$ in the language and elements $d_1,\ldots,d_n\in\DDD$, the map $\mu$ assigns $P(d_1,\ldots,d_n)$ either a $\text{true}$ or $\text{false}$ value.

From there, sentences in $\LLL\cup\DDD$ are assigned a truth value according to the rules \begin{equation} \begin{split} \neg F \text{ is true in } \MMM \ \ \ & \text{if and only if} \ \ \ \ F \text{ is not true in } \MMM; \\ F \land G \text{ is true in } \MMM \ \ \ & \text{if and only if} \ \ \ \ F \text{ is true in } \MMM \text{ and } G \text{ is true in } \MMM; \\ \exists x F(x) \text{ is true in } \MMM \ \ \ & \text{if and only if} \ \ \ \ F(d) \text{ is true for some } d\in\DDD. \end{split} \end{equation}

What do I mean when claiming the definition is equivalent?

I do not mean to say that any model (as it is usually defined) is of the form written above, nor that this alternative definition is 'philosophically' equivalent.

What I do mean is that concepts of satisfiability, tautology, contradiction, consistency, completeness, etc. carry over without change. For example, we can define a sentence $\phi$ as satisfiable by stating some model makes it true, and it matters not whether by 'model' we mean the usual definition or the one I wrote above.

Similarly, results such as the Model Existence Theorem remain true with this new definition.

If I'm understanding this post correctly, my proposal goes along the lines of the last section of the OP's, but assigning values to sentences in $\LLL\cup\DDD$ (as opposed to merely in $\LLL$) removes the need to speak of witnesses.

Am I wrong?

Sam
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1 Answers1

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Let $\mathcal{L}$ be the language with two constant symbols $c$ and $d$.

Let $\varphi$ be the $\mathcal{L}$-sentence $c = d$.

Under the ordinary definition of model, $\varphi$ is satisfiable. Take $M = \{*\}$, and interpret $c^M = *$ and $d^M = *$.

Under your definition of model, $\varphi$ is not satisfiable. Indeed, if the domain of $M$ is a superset of the constants in the language, $c$ and $d$ will not be equal in $M$, so $M$ will fail to satisfy $\varphi$.

In first-order logic without equality, this problem goes away, and I believe your proposal provides a satisfactory semantics for first-order logic without equality.

If you want to use implement your proposal in ordinary first-order logic (which includes equality), you'll need to fix this problem by introducing a non-standard way of interpreting equality.

Alex Kruckman
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  • Oh God, I completely missed that! +1. In first-order logic with equality, would my proposal give models isomorphic to models (as usually defined) that map different constants to different elements in the domain of discourse? – Sam Feb 04 '25 at 14:59
  • @Sam Yes, every model in which distinct constants are interpreted as distinct elements is isomorphic to one of your type. Just replace each element $c^M$ with the constant symbol $c$. – Alex Kruckman Feb 04 '25 at 15:01
  • One last question! If I was to define a model as a triple $(\mathcal{D}, \mathcal{R}, \mu)$ where $\mathcal{R}$ is an equivalence relation in $\mathcal{D}$, and set $a = b$ as true iff $a\mathcal{R}b$, would the definition now be 'fixed'? – Sam Feb 04 '25 at 15:12
  • @Sam Almost - you also need the other symbols in the language to respect $\mathcal{R}$. For example, if $E$ is a binary relation in the language, $a\mathcal{R}a'$, and $b\mathcal{R}b'$, then we should have $aEb$ if and only if $a'Eb'$. – Alex Kruckman Feb 04 '25 at 15:29