Maximal Consistent Sets of wff

Question

From 'A New Introduction to Modal Logic', p. 115. Theorem 6.3:

Suppose that Δ is an s-consistent set of wff. Then there is a maximal s-consistent set of wff Γ such that Δ ⊆ Γ.

The proof essentially consists in adding to the original s-consistent set (Δ), one step at a time, every wff (in Δ) or its negation. The resulting set (Γ) is not just s-consistent but s-consistent AND maximal (where every wff α in the set is either α or ~α).

Given the definition of s-consistency (Δ is s-consistent if there is no α1,..., αn ϵ Δ such that ⊢ ~(α1 Λ ... Λ αn) i.e. if no contradictions arise amongst the members of Δ), it seems to me that in every case the initial s-consistent set (Δ) would necessarily have to be maximal in the first place.

I struggle to imagine a case where the initial set (Δ) is s-consistent but not maximal. Could someone provide an example of an s-consistent but not maximal set? Aren't all s-consistent sets of wff necessarily maximal?

Thank you.

Answer 1

If we assume that $\varphi$ and $\varphi \rightarrow \psi$ are consistent formulas, then $\{\varphi, \varphi \rightarrow \psi\}$ is a consistent set, but it is clearly not maximal since e.g. $\psi \notin \{\varphi, \varphi \rightarrow \psi\}$.

Answer 2

Every maximal consistent set (m.c.s.) can be imagined as "one half" of the set of all formulas: for each formula $\alpha$, it contains either $\alpha$, or ${\sim}\alpha$, but not both, of course. Therefore, every m.c.s. must be infinite (since the set of all formulas is infinite).

Hence, if $\Delta$ is a finite consistent set of formulas, then $\Delta$ is not maximal. Of course, there are also infinite consistent sets that are still not maximal – e.g. any consistent normal modal logic; say, $K$.