# How to use nonlogical axioms in deduction

I'm a little bit unclear about the concept of nonlogical axioms and their potential usage in deduction.

According to this question, nonlogical axioms are those are which not universally valid and could not be presented by logical quantifiers and connectives.

Formal definition of deduction includes three seminal concepts: logical axioms, nonlogical axioms and rules of inference. If we admit that nonlogical axioms could not be stated in terms of logical concepts, how could we use them in deduction based inference rules?

• See e.g. first order Peano axioms : in order to prove theorem about arithmetic, we need logical axioms, like e.g. $\forall x (x=x)$, as well as "specific" arithmetical axioms, like e.g. $\forall n (S(n) \ne 0)$. – Mauro ALLEGRANZA Jan 18 '17 at 18:35
• "nonlogical axioms are those are which not universally valid" : CORRECT. – Mauro ALLEGRANZA Jan 18 '17 at 19:37
• "could not be presented by logical quantifiers and connectives" ? Of course, a non-logical axiom need some symbol in addition to quantifiers and connectives (and equality), like the binary predicates $\in$ (for set theory) and $<$ (for arithmetic). But we can use them also in universally valid sentences, like e.g. : $\forall x \forall y(x \in y) \lor \lnot \forall x \forall y (x \in y)$. – Mauro ALLEGRANZA Jan 18 '17 at 19:39
• Even if one defines $\le$ in the reference language of the axiom, its usage within the axiom labels it to a "nonlogical" one, right? – Roboticist Jan 18 '17 at 22:10

## 1 Answer

If you want to prove, say, a property of groups, your nonlogical axioms are going to be a suitable choice of axioms for group theory, which characterize associativity, existence of identity and inverses. The logical axioms, on the other hand, characterize sound reasoning, regardless of the intended application.

• So, every set of function and relation symbols, as a subset of all possible defined functions and relations corresponding to groups (as an instance ) represents nonlogical axioms?! – Roboticist Jan 18 '17 at 18:29
• No, axioms are (complete) sentences. For instance, if $\circ$ is the function symbol to be interpreted as the group operator, $\forall a \,. \forall b\,. \forall c \,. (a \circ b) \circ c = a \circ (b \circ c)$ is a nonlogical axiom. – Fabio Somenzi Jan 18 '17 at 18:37
• @Roboticist Maybe it's useful to point out that there are two distinct, albeit related, concepts: nonlogical symbols and nonlogical axioms. The constant, function, and relation symbols (and, according to some authors, also the quantifiers) are known as nonlogical symbols. Their meanings depend on their interpretations. Logical symbols (like $\wedge$) never change their meaning. – Fabio Somenzi Jan 18 '17 at 22:38
• Thanks, considerably for your explanations. – Roboticist Jan 18 '17 at 22:39