Re: H(D,D) cannot even be asked about the behavior of D(D) --- Boilerplate Reply

On 6/21/24 4:52 PM, olcott wrote:

On 6/21/2024 3:00 PM, Richard Damon wrote:

On 6/21/24 3:45 PM, olcott wrote:

On 6/21/2024 2:33 PM, Richard Damon wrote:

On 6/21/24 3:19 PM, olcott wrote:

int sum(int x, int y){ return x + y; }
When this program is asked: sum(3,4) this maps to 7.
When this program is asked: sum(5,6) this DOES NOT map to 7.

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Right.
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When H is asked H(D,D) this maps to D correctly simulated by H.
When H is asked H(D,D) this DOES NOT map to behavior that halts.
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Nope. H(M,d) is DEFINED (if it is correct) to determine if M(d) will Halt.
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If one "defines" that the input to H(D,D) maps to the behavior
of D(D) yet cannot show this because it does not actually
map to that behavior *THEN THE DEFINITION IS SIMPLY WRONG*

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But we CAN show that it maps to the behavior of D(D) (at least when the representation of D includes the H that is giving the 0 answer) by just runnig it and seeing what it does.
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 No you cannot show that the mapping for the input to
H(D,D) maps to the behavior of D(D).

The DEFINITION of a Halt Decider gives what H is SUPPOSED to do, if it is one.
You claim it is a correct Halt decider

 That you saw Bill's identical twin brother Jack rob
a liquor store does not prove that Bill robbed
that liquor store.
 Instead you simply assume that the mapping from a
non-input must equally apply.

Nope,

 You assume that the call to H(D,D) from D correctly
simulated by H *must return* against the verified facts
that it does not return.

No, I presume that a call from D to H(D,D) does what it was DEFINED to do, and that is to ask H to decide on the behvior of the computation D(D).
If that doesn't do that, then you are just admitting you have been lying for all these years.

 The directly executed D(D) is essentially the first call
in a recursive chain where the second call is always aborted.
*these two calls are not identical*

So, you admit that H isn't a pure functions, and not the equivalent to the needed Turing Machine,
Thus, you have wasted your last two decades.

 If H(D,D) did not correctly recognize that it must abort
the simulation of its input to correctly prevent its own
non-termination the directly executed D(D) would never halt.

But it does abort it simulation. YOU are the one confusing Bill with his twin brother Jack.
Which just shows you to be an INTENTILNAL liar, as your deception has been mentioned before, but you keep on repeating it.

 

Thus, the actual defined mapping exist, it just isn't comutable by H.
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 Misconceptions do not count as mapping.
int sum(int x, int y) { return x + y; }
Does not map sum(5,6) to 7.

Right, because that uses the wrong input

The input to H(D,D) does not map to halting.

It does if H is supposed to be a Halt Decider.

 H(D,D) is not free to simply assume that the call from
D(D) to H(D,D) will return.
 

No, it can't assume it, but it can't assume it isn't going to return either. You just lie to your programs as you lie to everything else.
Face it, you are just accumulating rebuttals that you just refuse to evern try to answer, because I think you know in your heart you ideas are just indefensable.
You are just establishing your lasting legacy of the man who would not stop lying, as he didn't know the basics of what he was talking about.
Maybe you should leave your brain to science to see if they can figure out what made you so unabe to learn definitions.