Re: Simulating termination analyzers by dummies --- criteria is met

Am Mon, 24 Jun 2024 16:10:00 -0500 schrieb olcott:

In my case people have been disagreeing with the semantics of the x86
programming language for three years when they have insisted that D
correctly simulated by H must have the same behavior as the directly
executed D(D).

What are the rules on how much a simulator can diverge from the actual
behaviour of its input?

When we stipulate that the only measure of a correct emulation is the
semantics of the x86 programming language then we see that when DDD is
correctly emulated by H0 that its call to H0(DDD) cannot possibly
return.

With that I agree. It follows that H0 does not simulate correctly.

When we define H1 as identical to H0 except that DDD does not call H1
then we see that when DDD is correctly emulated by H1 that its call to
H0(DDD) does return. This is the same behavior as the directly executed
DDD().

That is the actual behaviour. If a simulator does something different,
it is wrong. A simulation does not change behaviour.

--
Man kann mit dunklen Zahlen nicht rechnen. Für die eigentliche Mathematik
sind sie vollkommen nutzlos. --Wolfgang Mückenheim