On 2022-01-30 19:05, olcott wrote:
On 1/30/2022 7:45 PM, Richard Damon wrote:
On 1/30/22 7:21 PM, olcott wrote:
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
These statements need the conditions, that H^ goes to H^.Qy/H^.Qn iff
H goes to that corresponding state.
⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H in the
same way that (5,3) is syntactically specified as an input to Sum(5,3)
Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to embedded_H in the >> same way that (1,2) is NOT syntactically specified as an input to
Sum(5,3)
I promised myself I wouldn't involve myself in your nonsense any
further, but here you've made such a terribly inaccurate analogy that I thought I had to comment.
The inputs to a function such as SUM(X, Y) are two REPRESENTATIONS of integers. If SUM were a Turing Machine, these would be two strings in
the alphabet of the TM. if this were a C function, X and X would be
strings of bits which form the twos complement representation of some integer. In neither case would the inputs be actual, mathematical
integers. C might use the term 'integer' as one of its built in types,
but C integers are NOT elements of ℤ. They are REPRESENTATIONS of the supported subset of ℤ.
So ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H in the same sense that ⟨5⟩ ⟨3⟩ are
the inputs to SUM.
Ĥ ⟨Ĥ⟩ is not the input to embedded_H in the same sense that the actual mathematical integers 3 and 5 are not inputs to SUM.
If your going to make analogies, at least make ones that are accurate.
SUM takes REPRESENTATIONS of integers as its inputs, but it answers
about the ACTUAL integers described by those representations. To talk
about the sum of two representations is meaningless. Only actual
integers have sums.
In exactly the same way, embedded_H takes a REPRESENTATION of some TM ⟨Ĥ⟩ as part of its input but it answers about the ACTUAL TM described by that input, Ĥ.
To talk about whether a representation of a TM halts is
meaningless since only actual TMs, not representations of TMs, can halt.
The conditions which Richard indicates above (following Linz) are
therefore the correct ones.
In a previous post which I can't be botherered to find, you claimed that
when the input to embedded_H is ⟨Ĥ⟩ ⟨Ĥ⟩ that embedded_H can only be expected to answer about its actual inputs and not its 'enclosing TM'.
Yes, it must answer about its input, but if its input is ⟨Ĥ⟩ ⟨Ĥ⟩, then
BY THE DEFINITION OF A HALT DECIDER is must determine whether Ĥ applied
to ⟨Ĥ⟩ halts.
And that computation happens to be the EXACT SAME
computation as its 'enclosing TM'. So it is answering about *both*.
André
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