I'm a maths student, learning a bit about FOL. So if something I say doesn't make much sense, please point it out.
How uniquely determined are the semantics of First-Order logic when we fix the valid sentences of the language?
More specifically, can we change the recursive definition of satisfiability, 'meaningfully', without reducing the set of valid sentences, that is, change satisfiability in a way that the usual syntactical deductive system remains sound.
I noticed that we could define the semantics in a trivial way, by just defining that every sentence is satisfied by every model interpretation. This change would keep the usual tautologies because then, every sentence would be a tautology (in these trivial semantics). So to avoid this, let's say that the only requirement for the alternative semantics would be to keep the usual Negation as it is: M |= ¬ φ[v] iff M |# φ[v].
What led me to this question was related to consistency, so I was already expecting to not change the semantics of 'negation' as it would be necessary for the syntactical definition of consistency to work as intended.
I imagine that the usual tautologies, that need to remain tautologies, will fix the semantics of the boolean conectives. But I'm not sure this is the also the case for the quantifiers, so there might be room for change here.
I hope the question is somewhat clear, let me know if it needs more clarification.
Edit: I've read on the wikipedia page about Second-Order logic (in the section called Semantics) the following: "Unlike first-order logic, which has only one standard semantics,[...]". I'm not sure about what is being said here, maybe this means that the answer to my question is 'No, there is only one way of defining semantics with these requirements.'- If that's the case, I would still like to know why.
Edit 2: By different semantics I mean this: Let $S_1$ and $S_2$ be semantics for FOL. These semantics will be considered different iff there is some theory T and a structure M such that $S_1$ says M is a model of T, and $S_2$ says it isn't.