Location: EU Supported » PASCAL - Pattern Analysis, Statistical Modelling and Computational Learning » Summer Schools in Logic and Learning, Canberra 2009

Introduction to Modal Logic

author: Rajeev P. Goré, Australian National University
published: April 1, 2009,   recorded: January 2009,   views: 51125
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Description

We cover the syntax, Kripke semantics, correspondence theory and tableaux-style proof theory of propositional modal and temporal logics. These logics have important applications in a diverse range of fields incuding Artificial Intelligence, Theoretical Computer Science and Hybrid Systems.

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Reviews and comments:

Comment1 giuseppe motta, July 27, 2010 at 11:12 a.m.:

can you please show also lesson "5" of Rajeev Goré cours about modal logic?

Why is it not there?

That makes very diffcoult to follow part 6 too...

Thank you,

Giuseppe Motta

Comment2 Balaji, October 17, 2011 at 1:01 p.m.:

Just amazing. One of the best introductions to Modal Logic. I will all his class lectures were recorded and published online.

- Balaji

Comment3 Aglaophamus, November 30, 2013 at 2:20 a.m.:

How can (box phi) and (box not-phi) be true in the same world that doesn't have an immediate R-successor? Can anyone explain? I'm confused about this point.

Comment4 Lucifer, January 31, 2014 at 3:14 p.m.:

Aglaopgamus@

It owes to the fact that to make box-phi true, you must satify not-diamond-not-phi. But to do that, diamond-not-phi must be false. This means that diamond-not-phi must be false, i.e. that it ain't true.

Consider what makes it true: that some world accessible to the targeted world makes -phi true, i.e. that phi is false at some world relative to it. Since there are no such world, the sentence diamond not-phi is not possible, hence not-diamond-not-phi, hence box-phi.

The same goes for box-not-phi.

You can't make any possibility assertions true, though, as far as I can tell without thinking... don't know if I got it all wrong, though.

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