modal logic exercise
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From: Max on 16 Jan 2010 21:09 I just started going through _Dynamic Epistemic Logic_ by van Ditmarsch et al. I am a little rusty on modal logic and have gotten stuck on an early exercise. The problem is actually misstated in the book, but I'm pretty sure that what they mean (or, the component of what they're asking for that I'm stuck on) is: Let the logic _K''_ result from the usual logic _K_ by replacing the axiom scheme k: L(A->B) -> (LA -> LB) with the scheme k'': L(A.B) -> LA . LB. Show that every instance of the k scheme is derivable in the logic _K''_. Can somebody please help me out! Thanks, Max
From: Frederick Williams on 17 Jan 2010 10:38 Max wrote: > > I just started going through _Dynamic Epistemic Logic_ by van > Ditmarsch et al. I am a little rusty on modal logic and have gotten > stuck on an early exercise. > > The problem is actually misstated in the book, but I'm pretty sure > that what they mean (or, the component of what they're asking for that > I'm stuck on) is: > > Let the logic _K''_ result from the usual logic _K_ by replacing the > axiom scheme > > k: L(A->B) -> (LA -> LB) > > with the scheme > > k'': L(A.B) -> LA . LB. > > Show that every instance of the k scheme is derivable in the logic > _K''_. > > Can somebody please help me out! It helps to recall that A->B is A<->A.B and to derive the rule A<->B / LA<->LB. -- Pigeons were widely suspected of secret intercourse with the enemy; counter-measures included the use of British birds of prey to intercept suspicious pigeons in mid-air. Christopher Andrew, 'Defence of the Realm', Allen Lane
From: Max Weiss on 17 Jan 2010 16:59 On Jan 17, 7:38 am, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > It helps to recall that A->B is A<->A.B and to derive the rule > A<->B / LA<->LB. Thanks! Could you please spell out the derivation? ---Max > Max wrote: > > > I just started going through _Dynamic Epistemic Logic_ by van > > Ditmarsch et al. I am a little rusty on modal logic and have gotten > > stuck on an early exercise. > > > The problem is actually misstated in the book, but I'm pretty sure > > that what they mean (or, the component of what they're asking for that > > I'm stuck on) is: > > > Let the logic _K''_ result from the usual logic _K_ by replacing the > > axiom scheme > > > k: L(A->B) -> (LA -> LB) > > > with the scheme > > > k'': L(A.B) -> LA . LB. > > > Show that every instance of the k scheme is derivable in the logic > > _K''_. > > > Can somebody please help me out! >
From: William Elliot on 18 Jan 2010 06:16 On Sun, 17 Jan 2010, Frederick Williams wrote: >> Let the logic _K''_ result from the usual logic _K_ by replacing the >> axiom scheme >> >> k: L(A->B) -> (LA -> LB) >> >> with the scheme >> >> k'': L(A.B) -> LA . LB. >> >> Show that every instance of the k scheme is derivable in the logic >> _K''_. >> >> Can somebody please help me out! > It helps to recall that A->B is A<->A.B and to derive the rule > A<->B / LA<->LB. That can be derived? Then (A <-> B) -> (LA <-> LB). L(A <-> B) -> (LA <-> LB) is believable as is |- A <-> B implies |- LA <-> LB.
From: David C. Ullrich on 20 Jan 2010 14:12
On Mon, 18 Jan 2010 03:16:41 -0800, William Elliot <marsh(a)rdrop.remove.com> wrote: >On Sun, 17 Jan 2010, Frederick Williams wrote: > >>> Let the logic _K''_ result from the usual logic _K_ by replacing the >>> axiom scheme >>> >>> k: L(A->B) -> (LA -> LB) >>> >>> with the scheme >>> >>> k'': L(A.B) -> LA . LB. >>> >>> Show that every instance of the k scheme is derivable in the logic >>> _K''_. >>> >>> Can somebody please help me out! > >> It helps to recall that A->B is A<->A.B and to derive the rule >> A<->B / LA<->LB. > >That can be derived? >Then (A <-> B) -> (LA <-> LB). Why? >L(A <-> B) -> (LA <-> LB) is believable as is > >|- A <-> B implies |- LA <-> LB. Are you certain you know what he meant by the symbol "/"? I'm not, but I _suspect_ that when he says "A<->B / LA<->LB" that means exactly the same thing as "|- A <-> B implies |- LA <-> LB". > |