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Does anyone know of any propositions that would suffice, along with the necessitation rule, to prove either of the following two propositions?

$\Box(p \to q) \to (\Box p \to \Box q)$

$(\Box p \land \Box q) \to \Box (p \land q)$

I know the first is usually taken as axiomatic in modal logic, but I'm trying to find other propositions that, together with necessitation (that all theorems of propositional logic are necessary truths), can prove the two propositions above.

Tsvi Benschar
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Obviously you can trivially prove the distribution axiom ($\Box(p \to q) \to (\Box p \to \Box q)$) in the traditional C.I Lewis's axiomatic system K, but not from the necessitation with other propositions, otherwise it's not required to have this as an axiom.

Of course you can use a natural deduction system K to prove both of your above propositions easily, even the necessitation rule itself can be derived from such a ND system which usually has the $\Box$E and $\Box$ In modal rules in addition to the usual PL rules as referenced here:

Let us take as an example the ND formalization...; for simplicity we restrict considerations to rules for $\Box$ (necessity). ($\Box$ E) is obvious: $\Box \varphi \vdash \varphi$. With ($\Box$ I) the situation is more complicated since it is based on the following principle:

If $\varphi_1, ..., \varphi_n \vdash \psi$, then $\Box\varphi_1, ..., \Box\varphi_n \vdash \Box\psi$

Right after this quoted section you'll see a typical ND proof of your first proposition without any axioms or even the necessitation involved.

To prove your 2nd proposition under such system is also easily derived as a single step per above $\Box$ I rule:

$$\frac{p, q \vdash p \land q}{\Box p, \Box q \vdash \Box (p \land q)}$$

Or alternatively you can use a Fitch style subproof similar to the reference proving your 1st proposition...

cinch
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  • I'm looking ideally for other propositions that alone, with maybe the necessitation rule, can show one of the above. Natural deduction works but the $\Box$E and $\Box$I are essentially just allowing for the natural deduction analog to the K axiom. I'm looking for something that's conceptually different. The reason is because I'm working in a certain weird non-normal modal context where either of the above leads to a contradiction in the semantics, so I'm looking for other things that imply them so I know what else the semantics prohibit. – Tsvi Benschar Dec 11 '21 at 00:50
  • @TsviBenschar That sounds a rather stringent proprietary requirement... Can you share more about your context reference or book name? I know it may not be a complete book otherwise it will include all relevant rules already usually, so is it an open research project? – cinch Dec 11 '21 at 01:14
  • Its for a philosophy term paper im doing. If you define necessitation so that $\Box p$ iff $p$ has some sufficiently high probability, then having $\Box p \land \Box q$ does not always imply $\Box(p \land q)$, since the probability of conjunctions might be lower than of their conjuncts. So the second proposition I listed isnt allowed. The K axiom and $\Box(p \to (q \to (p \land q)))$ imply the second proposition, so the K axiom isn't allowed either. I'm trying to figure out if there's other stuff I can prohibit. Necessitation is fair game because the probabilities of theorems are always 1. – Tsvi Benschar Dec 11 '21 at 01:22
  • @TsviBenschar hmm sounds totally different than your original posted question... I need some time to tinker around... In the meantime since it's a phil paper, maybe you can change your question given above context and post on PhilosophySE since more philosophers are interested and had thought about such logics? – cinch Dec 11 '21 at 01:36
  • @TsviBenschar Also I'm not convinced re your critical assumption "the probability of conjunctions might be lower than of their conjuncts". Can you give a concrete example to show how independent random variable's probability's multiplication principle will be broken under modality? – cinch Dec 11 '21 at 01:55
  • Say your threshold for belief is some rational number $t=\frac{p}{q}$ where $0<p<q$. Take two random number generators, set their range to be the integers $[1,q]$, and have them each pick $p$ numbers randomly, so each number has an equal chance of being picked. Take an $i,j$ from $[1,q]$. $i$ has $t$ probability of being picked by the first generator by construction, and likewise $j$ by the second generator. But the conjunction of "$i$ is picked by the first and $j$ by the second" is $t^2$, since the generators are presumably independent, which is less than $t$. – Tsvi Benschar Dec 11 '21 at 02:07
  • @TsviBenschar mix probability with modal logic needs to modify semantics first like a recent paper here which employs a probabilistic Kripke model. As for syntactic rules it needs to add something like If X is a formula, 0 < r ≤ 1, and α ∈ {<, =} then (K(X) α r) is a formula. So as a basic idea you need to formulate propositions differently, but the □E and □I are essential otherwise it cannot be related to modal logic. The art is to decide the value of r for each such formula... – cinch Dec 11 '21 at 02:50
  • Aren't there other modal logics that reject the K axiom? It requires non standard semantics, but why is this unrelatable to modal logic? – Tsvi Benschar Dec 11 '21 at 03:52
  • Let us continue this discussion in chat. – cinch Dec 11 '21 at 04:01