Sujet : Re: Two dozen people were simply wrong --- Try to prove otherwise --- pinned down
De : noreply (at) *nospam* example.com (joes)
Groupes : comp.theoryDate : 02. Jun 2024, 11:18:35
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <v3hddb$2psm0$1@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
Am Sat, 01 Jun 2024 13:44:21 -0500 schrieb olcott:
On 6/1/2024 1:40 PM, Fred. Zwarts wrote:
Op 01.jun.2024 om 18:24 schreef olcott:
On 6/1/2024 11:19 AM, Fred. Zwarts wrote:
Op 01.jun.2024 om 18:13 schreef olcott:
On 6/1/2024 10:56 AM, Richard Damon wrote:
On 6/1/24 11:30 AM, olcott wrote:
>
>
HH correctly reports that because DD calls HH(DD,DD) in recursive
simulation that DD never halts.
>
HHH(HH,DD,DD) would report that HH halts.
>
>
Maybe. And H1 (DD,DD) would report that DD halts.
In the recursive simulation by HH, neither the simulation of DD, nor
the simulation of HH halts. If one of them would halt, the other one
would halt as well.
So HH 'correctly' reports that both DD and HH do not halt, because they
both keep starting an instance of each other.
In case you didn't know pure functions must halt because they must
return a value.
That means it terminates/returns, right? Then H is either not „pure”,
or execution proceeds past line 4.
-- joes
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