Related articles |
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help with basic derivations jon.gallagher.04@gmail.com (jon.gallagher.04) (2009-01-26) |
Re: help with basic derivations m.helvensteijn@gmail.com (Michiel Helvensteijn) (2009-01-28) |
Re: help with basic derivations jon.gallagher.04@gmail.com (jon.gallagher.04) (2009-01-30) |
Re: help with basic derivations m.helvensteijn@gmail.com (2009-02-03) |
From: | Michiel Helvensteijn <m.helvensteijn@gmail.com> |
Newsgroups: | comp.compilers |
Date: | Wed, 28 Jan 2009 00:28:55 +0100 |
Organization: | Compilers Central |
Keywords: | types |
Posted-Date: | 27 Jan 2009 19:09:32 EST |
jon.gallagher.04 wrote:
> The text I am using states that this evaluation rule means that when
> t1 = true, we can be sure the rule will always evaluate to t2. I
> believe it. I can't figure out how to prove it, or even how a
> derivation would get there. I try the following
>
> t1 <- T t2 <- T t3 <- T t1 = true
> ---------------------------------------------------
> if t1 then t2 else t3 t1 = true
> ------------------------------------------
> if true then t2 else t3 ---> t2
>
> From here how do I conclude t2?
>
> Thank you for any help you can give!
What you're describing looks like operational semantics for a simple
programming language. I'm currently working on a similar set of operational
semantic rules for another project.
I believe that you don't have to prove that "if true then t2 else t3"
evaluates to t2. I believe that this evaluation rule exists to define the
semantics of the if-then-else construct. It is not something for you to
prove. It is a basic building block you can use to prove more complicated
derivations.
--
Michiel Helvensteijn
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