On 8/1/2025 7:10 PM, Ben Bacarisse wrote:Proven what? And why can't halt deciders be decided?
Mr Flibble <flibble@red-dwarf.jmc.corp> writes:
The two types of programs are:
(A) halt deciders;
(B) programs being proven by the decider.
WDYM "encode"?Type constraints:
(B) cannot encode (A) due to uni-directionality.
*There is a violated type constraint in the Linz proof*Why should it? It is just a constraint expressed using variables.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.
The "if" statements are incorrect because Ĥ is not of the finite string type.
The Linz proof requires Ĥ.embedded_H to report on its own behaviorAren't those the same?
instead of the behavior of its input.
On 8/2/2025 1:47 PM, joes wrote:
Am Fri, 01 Aug 2025 19:21:39 -0500 schrieb olcott:
On 8/1/2025 7:10 PM, Ben Bacarisse wrote:Proven what? And why can't halt deciders be decided?
Mr Flibble <flibble@red-dwarf.jmc.corp> writes:
The two types of programs are:
(A) halt deciders;
(B) programs being proven by the decider.
WDYM "encode"?Type constraints:
(B) cannot encode (A) due to uni-directionality.
*There is a violated type constraint in the Linz proof*Why should it? It is just a constraint expressed using variables.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.
The "if" statements are incorrect because Ĥ is not of the finite string >>> type.
Strings can't be applied or run.
The Linz proof requires Ĥ.embedded_H to report on its own behaviorAren't those the same?
instead of the behavior of its input.
No Turing machine can even directly look at another
Turing machine because no TM takes another actual TM as input.
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