Sujet : Re: D correctly simulated by H cannot possibly halt --- templates and infinite sets
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theory sci.logicDate : 30. May 2024, 02:02:32
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <v38j78$2foi0$13@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13
User-Agent : Mozilla Thunderbird
On 5/29/24 8:53 PM, olcott wrote:
On 5/29/2024 7:47 PM, Richard Damon wrote:
On 5/29/24 8:21 PM, olcott wrote:
On 5/29/2024 7:09 PM, Richard Damon wrote:
On 5/29/24 8:01 PM, olcott wrote:
On 5/29/2024 6:47 PM, Richard Damon wrote:
*Formalizing the Linz Proof structure*
∃H ∈ Turing_Machines
∀x ∈ Turing_Machines_Descriptions
∀y ∈ Finite_Strings
such that H(x,y) = Halts(x,y)
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And since NO H, can get right the H^ built to contradict IT, that claim is proven false.
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YOU KEEP TRYING TO GET AWAY WITH CHANGING THE SUBJECT
THE ABOVE FORMALIZATION IS CORRECT
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How?
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The above is the question that Linz asks and the he gets
an answer of no, no such H exists.
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So, you now agree with Linz. Good.
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I said that Linz says that. The point is that the Linz
template examines an infinite set of Turing Machine / input
pairs the same way my H/D template references an infinite set
of C function / input pairs.
The difference is, In Linz's formulation, each machine is INDIVIDUALLY EVALUTED with its inputs, and a specific input is shown to exist for each one that makes it wrong.
YOU try o merge them all together to claim that since none of them reach a final state in its partial simulation (for those that answer) that means that all the inputs represented non-halting machines,
THAT is just invalid logic.
You just don't understand how to do logic with qualifiers.