0

I use a natural deduction calculus for modal propositional logic, but my question eventually is about any (sound) modal calculi with/without axioms.

Just as an example take a proof like $\square$A $\vdash$ A that any calculus based on T or stronger can prove.

Now, $\square$A $\vdash$ A looks like a false proof because in modal logic you just cannot conclude from some $\square$A to some A independent from the world they are in, but only if they play both in the same world. So are proofs like $\square$A $\vdash$ A just a short-cut for $w_{i}\square$A $\vdash$ $w_{i}$A, i.e. in your mind you always have to add $w_{i}$ before any wff?

God
  • 95
  • See Possible Worlds Semantics for Modal Logic – Mauro ALLEGRANZA Dec 01 '23 at 06:43
  • But my question concerns not the semantics but the syntactics. Again, I want to know how a proof $\square$A $\vdash$ A ought to be read: like I suggest with reference to an arbitrary but fixed world $w_{i}$ or not; but then as I wrote I‘d find it irritating because $\square$A $\vdash$ A cannot mean that from any $\square$A you can conclude A because e.g. when $\square$A is true in world w1 then A could still be false in w2 (assuming a T-model for instance). – God Dec 02 '23 at 00:52
  • Syntactical prrof means axioms+rules. – Mauro ALLEGRANZA Dec 02 '23 at 07:53
  • I think I got it myself. In a modal calculus A $\vdash$ B (A, B are wff of a modal logic language) does not care about worlds, it just means that B follows from A, period. If the modal calculus wants to be sound it must also hold A $\vDash$ B and because of the possible world semantics that automatically translates into A,w $\vDash$ B,w. – God Dec 02 '23 at 14:18

0 Answers0