PA theorem
Mario Carneiro
di.gama at gmail.com
Sat Jul 17 23:03:59 EDT 2021
Yes. You can prove that by induction on z:
If P(z), then we are done by (ii). Otherwise, if z = 0, then by (i) the
P(x) that exists is > 0 because it is not z, and if z = w+1 then there
exists P(y) with y>w by the inductive hypothesis, which is not z, so in
fact y>z.
Mario Carneiro
On Sat, Jul 17, 2021 at 10:21 PM Giovanni Lagnese <giov.lagn at gmail.com>
wrote:
> It is true that, for any formula with a free variable P(x) in the language
> of first order arithmetic, it is a PA theorem that:
> If:
> i) an x exists such that P(x)
> ii) for each x such that P(x) there exists a y>x such that P(y)
> Then for every z there exists a k such that P(k) and k>z?
> Is this, for any formula with a free variable P(x), a theorem of PA?
>
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