Hi everyone,

I uploaded yesterday a preprint to the arXiv titled "Independence questions 
in a finite axiom-schematization of first-order logic" 
(https://arxiv.org/abs/2202.10383).

Although the axiomatization studied in that paper differs slightly from 
that of set.mm, when applied to the set.mm axiomatization of first-order 
logic, setmm := {ax-mp, ax-gen, ax-1 to ax-13}, it proves:
 - independence of ax-11 in setmm \ {ax-12, ax-13},
 - independence of ax-13 in setmm \ {ax-12},
and adapts classical results to prove:
 - independence of ax-mp, ax-gen, ax-1 to ax-9, ax-12 in setmm,
 - independence of ax-10 in setmm \ {ax-5, ax-12}.
There are a few other results, and also an introductory section presenting 
the formal systems used, which may be useful as a companion introduction to 
Metamath for set.mm-like databases (its content will not be new to regular 
contributors of this mailing list).

The main new ingredient to prove partial independence of ax-11 and ax-13, 
which I called supertruth and discovered in August 2020, modifies the set 
of instances that a given scheme may have, hence can prove independence of 
schemes which are redundant at the object-level (see the paper for the 
definition).  I rapidly let Norm and Mario know about this, and that 
discovery slowly extended, thanks to their comments and over a bit more 
than a year, to the present paper, of which Norm knew the almost-final 
form.  The paper is dedicated to his memory.  I also extend my thanks here 
to Mario, whose comments were very useful.  I also mention Wolf in the 
acknlowledgments, since his work on reducing axiom dependencies was also 
useful to clear things up.

Of course, your comments are welcome.

Best regards,
BenoƮt

-- 
You received this message because you are subscribed to the Google Groups 
"Metamath" group.
To unsubscribe from this group and stop receiving emails from it, send an email 
to [email protected].
To view this discussion on the web visit 
https://groups.google.com/d/msgid/metamath/5ed340b6-22d7-4e13-8406-503121cff958n%40googlegroups.com.

Reply via email to