Why is Normality an axiom in Modal Logic?

Question

Why are we having $\square(A → B) → (\square A → \square B)$ as an axiom when we can prove* that if we have $\square(A → B)$ then we will have $(\square A → \square B)$?

*$B$ is true in every world by assumption which makes $\square B$ true and hence $\mathfrak{M}, w \nVdash \square A $ or $\mathfrak{M}, w \Vdash \square B$ for every $w$ in $\mathfrak{M}$ that is to say $\mathfrak{M}, w \Vdash \square A \rightarrow \square B$.

I am almost certain that I am completely wrong but I cannot find my mistake in my thinking.

Help, please.

Answer 1

You are conflating the proof theory of modal logic with the semantics of modal logic. We can approach verifying statements in two ways: 1) formal proofs from the axioms. 2) informal arguments for their validity in terms of Kripke frames. (Which is not to say these are the only two approaches, just the two relevant to this question.)

The theorems of system K (the smallest normal modal logic) can correspondingly be thought of in two ways. The first way is that a statement holds in K iff it can be formally proved by applying the rules/axioms of propositional logic, combined with the necessitation rule and the normality axiom. The second way is that a statement holds in K iff it is valid in all Kripke frames. The hard work is showing that these two sets of statements are actually the same. This is an example of a soundness/completeness result: the statements provable in some formal system are exactly those valid with respect to some semantics.

So given that fact, it is completely expected that axiom K is valid in all Kripke frames, cause obviously it can be proved in a formal system that takes it as an axiom. But it is in no way superfluous, because without it you cannot prove (using the proof theory approach) every statement that is valid in all Kripke frames.

The next normal modal logic people usually learn is system T, which is an extension of K with more valid statements. This extension can be thought of in two ways: 1) adding the axiom $\square \phi\to \phi$ to the formal system. 2) No longer requiring a valid statement to hold in all Kripke frames, rather just the reflexive frames. So in the first case you get more valid statements since you have another axiom to prove things with, and in the second you get more because you require the statement to hold in a smaller class of frames. And it turns out that either way results in the same set of valid statements: another soundness/completeness result.

One other thing to note: they are less well-known, but there are non-normal modal logics that don't admit the normality axiom. Of course, these won't have semantics simply in terms of Kripke frames like normal modal logics.

Answer 2

This looks to me like the distinction between Semantics and Syntax, as applied to modal logic.

What we want to do with formal logic is allow the manipulation of strings of symbols according to mechanical rules - the Syntax - to allow us to learn about logical facts about certain mathematical structures - the Semantics. From the point of view of the Semantics, the symbols we use have particular meanings that we all agree on, so that strings of symbols say things about mathematical structures. From the point of view of the Syntax, the symbols just start out as empty labels: they only gain their meanings because they obey certain mechanical rules that match what we expect those meanings to do.

For instance...

• Semantically, $\rightarrow$ means "Implies". Syntactically, we only start out with $\rightarrow$ being some connective of a pair of terms, and have to explicitly define Modus Ponens as a rule in order to make $\rightarrow$ act like "Implies".
• Semantically $\forall$ means "For All". In order to make the symbol $\forall$ act like "For All" syntactically, we have to define a rule that says that whenever we have $\forall x: P(x)$ we can substitute in any term $k$ to conclude $P(k)$.

In both these cases, we have to add those rules to our logical system in order to make the symbols act like we want them to.

You've shown that any symbol that means "Necessary" has to obey that rule, semantically. But syntactically, $\square$ just starts out as an empty symbol with no meaning behind it. If we didn't state our rules for using it, we wouldn't be able to do anything with it. That's why we have to state rules for using $\square$, one of which is $\square(A → B) → (\square A → \square B)$.

Answer 3

There are Natural Deduction systems which allow you to prove the Normality Axiom in a similar fashion as the semantic proof you gave. I’ll show you how it works, but note that this doesn’t work for an axiomatic system.

1.□(P->Q) (Assumption)

2.□P (Assumption)

3. □

4. P->Q (1,□-Reit.)

5. P (2,□-Reit.)

6. Q (4,5 M.P.)

7.□Q (3-6,□-Intro.)

As you can see, this proof uses a special box sub-proof in which only formulas under the scope of a box may appear. You can also use a diamond sub-proof in which a specific formula under the scope of a diamond may appear, as well as any formula under the scope of a box. This is unnecessary, as the box-subproof technique is sufficient for system K.