Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [modal-logic]

Questions relating to deductions relating to the expressions "it is necessary that" and "it is possible that"

Modal logic is an extension of propositional and predicate logic that expresses modalities, which are qualifications to a statement. The most commonly used modalities in mathematics are "possibly," "necessary," and "impossibly."

For more information, see these links:

566 questions
6
votes
1 answer

Prove that $\Diamond\Box p \rightarrow \Diamond (\Box (p\land q) \lor \Box(p\land\neg q))$ does not define a first-order condition on frames

First of all, the modal logic we are working with in this case is the basic one: that is, all propositional formulas, plus formulas of the form $\Diamond\phi$, where $\phi$ is any modal formula (we define $\Box\phi$ as $\neg\Diamond\neg\phi$). Let's…
hcp
  • 397
4
votes
2 answers

Equivalence between classical modal logic and propositional logic

I have read in some lecture notes that "pure classical modal logic without any additional conditions is nothing more than propositional logic". I cannot really see what this means and unfortunately the context is not helpful. Any guess? Here is the…
Mijito
  • 235
3
votes
1 answer

Is there an unsatifiable formula in modal logic K whose negation is not valid?

I just stumbled across this. In classic logic, the negation of a valid formula is unsatisfiable and vice versa. Given the usual Kripke semantics definitions of modal logic K (see below), this law seems only to hold in the former direction. But I…
smirk
  • 33
3
votes
1 answer

Proof that Scheme T implies reflexivity

In my modal logic book it's written that, for each frame $F(S,R)$ the accessibility relation $R$ is reflexive IF AND ONLY IF the scheme T:$\square A \implies A$ is valid in $F$. Even if I can easily prove that reflexivity $\implies$ T, I can't prove…
Giacomo Tagliabue
  • 165
  • 3
2
votes
1 answer

Can it be the case $p$,$\lnot p$ are true at $\mathbf w$ And $\mathbf w'$ respectively with $\mathbf w$ and $\mathbf w'$ have access to each other?

Given a model $\mathbf{M} = (\mathbf{W}, \mathbf{R}, \mathbf{V})$ for a set of atomic formulae $\Omega$. We have possible worlds $\mathbf{w}, \mathbf{w'} \in \mathbf{W}$, access relation satisfies $(\mathbf{w}, \mathbf{w'}),(\mathbf{w'},…
Bender
  • 57
  • 6
2
votes
1 answer

proving $\Box(p \to q) \to (\Box p \to \Box q)$ or $(\Box p \land \Box q) \to \Box (p \land q)$ from necessitation and other propositions

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…
Tsvi Benschar
  • 101
2
votes
0 answers

Show that the canonical model for $\mathbf{S5}$ is not universal

Show that the canonical model for $\mathbf{S5}$ is not universal. We know that the canonical model $M$ for $\mathbf{S5}$ is based on the class of frames $\mathscr{F}$ where $R$ is an equivalence relation. And since every universal relation is an…
beachtrees
  • 209
2
votes
1 answer

What kind of Kripke frame validates $(\Diamond p \wedge \Diamond q) \rightarrow \Diamond(p\wedge q)$?

$(\Box p \wedge \Box q) \rightarrow \Box(p\wedge q)$ is valid in K. But $(\Diamond p \wedge \Diamond q) \rightarrow \Diamond(p\wedge q)$ is not. I'm not sure what a frame that validates $(\Diamond p \wedge \Diamond q) \rightarrow \Diamond(p\wedge…
eyet
  • 23
2
votes
1 answer

Distribution axiom (K)

What is the reason for taking the following as the axiom in normal modal logics: □(p → q) → (□p → □q)? What is special about this formula that makes it an axiom? Does it express some kind of tautology? For example, the following is an axiom of…
user758536
2
votes
1 answer

$\sf K$-proof of $(\Box p \land \Diamond q) \rightarrow \Diamond (p\land q)$

$\sf K$-proof of $(\Box p \land \Diamond q) \rightarrow \Diamond (p\land q)$ I am stuck on the proof of this using the axiom system $\sf K$. This is the exercise 1.6.1 in Blackburn's modal logic. Any help, please? Thank you.
Y.X.
  • 3,995
2
votes
1 answer

modal logic proof question

If we have $$ \vdash A \rightarrow B $$then we can derive $$ \vdash \Box A \rightarrow \Box B $$ By necessity But we should then also be able to derive $$ \vdash \diamond A \rightarrow \diamond B $$ Intuitivly, it makes sense, but I am not sure how…
SilverTear
  • 201
1
vote
2 answers

Basic Modal Logic question #1

Its about two weeks I have started Cresswell's "A New Introduction To modal Logic". Now I've got a few questions on the text and I would deeply thank you if you help me clarify on…
Ak9
  • 571
  • 2
  • 10
1
vote
1 answer

Incomplete normal modal logic systems

Apart from the classical example of KH, given by axiom $\Box(\Box p\leftrightarrow p)\to \Box p$, are there any other examples of incomplete propositional normal modal logic systems defined by axioms in 1 variable and of modal degree $\le 2$? If so,…
JuneA
  • 356
1
vote
0 answers

Does Hennessy-Milner Theorem hold when I weaken its condition (image-finite) a little?

Hennessy-Milner Theorem says that For two image-finite models M,N, we have that the pointed models M,w and N,v are equivalent in semantics (all holds on M,w holds on N,v and vice versa) iff M,w and N,v are bisimilar. Then I wonder whether the…
Cleanlee
  • 11
  • 1
1
vote
0 answers

How many modal logics are there?

How many propositional modal logics exist? Is it a finite number, countably infinite, or continuum-sized? Of course, to answer this question, we would need a definition of a propositional modal logic.
user107952
  • 20,508
1
2 3 4 5