Sujet : Re: Simulating termination analyzers by dummies --- What does halting mean?
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theory sci.logicDate : 20. Jun 2024, 02:24:01
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <v4vsr1$ggem$3@i2pn2.org>
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User-Agent : Mozilla Thunderbird
On 6/19/24 1:09 PM, olcott wrote:
On 6/19/2024 11:43 AM, joes wrote:
Am Wed, 19 Jun 2024 09:23:04 -0500 schrieb olcott:
On 6/19/2024 6:30 AM, Richard Damon wrote:
On 6/18/24 10:51 PM, olcott wrote:
>
Thus according to your faulty reasoning when the source-code of a C
program is simulated by interpreter this is mere nonsense gibberish
having nothing to do what the behavior that this source-code
specifies.
IF you can show that a given simulation will produce the exact same
results as the direct execution of the program, then the simulation
will show the actual behavior of the program.
Now, if it doesn't. then it is gibberish.
No matter how much you try to simply ignore the verified fact that the
pathological relationship between an input and its termination analyzer
changes the behavior of this emulated input relative to the behavior of
its direct execution THIS VERIFIED FACT WILL NOT GO AWAY.
That's just wrong. A program/machine has a fixed behaviour.
>
Ignoring my conclusive proof that DDD correctly simulated by HH0
has different behavior that DDD() is simply deception.
You mean your lie that H0 even correctly simulated the input?
SInce the trace it gives didn't go into H0, it is proven to be an incorrect trace, and that you are just ignorant of the actual behavior of the machines you are talking about.
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