# Nonlogical axioms and their validity

I am reading A Friendly Introduction to Mathematical Logic by Leary and Kristansen. The authors show that the logical axioms are valid. Since the logical axioms are independent of the language, they are true in any $$\mathcal{L}$$-structure and any first order language $$\mathcal{L}$$.

This gets confusing when it comes to nonlogical axioms. Peano Axioms $$N$$ (see Example 2.8.3) are nonlogical axioms in the language of number theory $$\mathcal{L}_{NT}$$. Here's my question: If $$\mathfrak{N}$$ is the standard structure of the number theory, is it correct to say that $$\mathfrak{N}\models N$$? Is it simply true by definition? Or do we accept nonlogical axioms to be true in any structure?

Likewise, if we are in the language of set theory, do we call the structure $$\mathfrak{S}$$ to be "standard" where ZF is accepted as nonlogical axioms?

Also, the authors prove that there is no way that a set of $$\mathcal{L}_{NT}$$ sentence can ever characterise the standard structure, i.e., even if there was some other structure such that $$\mathfrak{A}$$ where the Peano axioms $$N$$ was true then it would not hold that $$\mathfrak{A}\simeq\mathfrak{N}$$.

• Exactly; we assume that Peano Axioms are true in the "standard model". And no; non-logical axioms are not valid i.e. true in very structure. The axiom $\lnot \exists x (0=s(x))$ does not hold in a domain with only one objcet. Commented Mar 21, 2021 at 7:55
• Your comment makes sense to me. There is no standard model for vector spaces. Speaking mathematically, we would not really call those models which do not satisfy the non logical axioms of vector space to really be a vector space. So, is there any point of studying those models which really do not satisfy the nonlogical axioms which "actually" define those algebraic structures?
– ashK
Commented Mar 21, 2021 at 8:20

If $$\mathfrak N$$ is the standard structure of the number theory, is it correct to say that $$\mathfrak {N} \vDash \mathsf {PA}$$ [where $$\mathsf {PA}$$ are the Peano Axioms]?
And no; non-logical axioms are not valid, i.e. true in every structure. The axiom $$¬∃x(0=s(x))$$ does not hold in a domain with only one object.