Text used in the video
https://youtu.be/UG-evnE3-p0
==== Regularity statement ====
https://en.wikipedia.org/wiki/Axiom_of_regularity
(Classically)
$\forall s.\Big(s\neq\emptyset,\to,\exists(x\in s). x\cap s=\emptyset\Big)$
==== History ====
https://en.wikipedia.org/wiki/Zermelo_set_theory
(No regularity or replacement)
Excerpt from: Thoralf Skolem - "Einige Bemerkungen zur axiomatischen Begründung der Mengenlehre" (1922)
A.k.a. "Some remarks on axiomatized set theory"
Also reprinted in a translation in "From Frege to Gödel", 1999, pp. 290-.
1925: von Neumann shills regularity for his ordinals.
==== Least number principle ====
==== von Neumann model ====
https://en.wikipedia.org/wiki/Set-theoretic_definition_of_natural_numbers
Set theoretical Neumann model for PA:
$0$ ... $\emptyset$
$Sx$ ... $x\cup{x}$
So
$1=S0={0}={\emptyset}$
$2=S1={0,1}={\emptyset,{\emptyset}}$
$3=S2={0,1,2}={\emptyset,{\emptyset},{{\emptyset}}}$
$4=S3={0,1,2,3}={\emptyset,{\emptyset},{{\emptyset}},{{{\emptyset}}}}$
$\dots$
$\omega={0,1,2,\dots}$
The Model proves all the axioms of the FOL theory.
$x < y$ ... $x\in y$
E.g.
$2<4\iff{\emptyset}\in{\emptyset,{\emptyset},{{\emptyset}},{{{\emptyset}}}}$
Note:
$n={k\in\omega\mid k < n}$
==== A world without regularity ====
$c={c}$
$c\in c$
(So $c < c$ if we want to use the notation $<$ also for other sets)
$Sc=c\cup{c}=c\cup c=c$
==== As statement in set theory ====
https://en.wikipedia.org/wiki/Taxicab_number
For any property $T$, if any number has the property, then there is a smallest natural with that property.
For a von Neumann natural
$n={k\in\omega\mid k < n}$
$\Theta$ ... ${n\in\omega\mid T(n)}$
$n\in\Theta$ ... $T(n)$
(Classically)
$\Theta\neq\emptyset,\to,\exists (n\in \Theta). n\cap \Theta=\emptyset$
$\exists m.T(m),\to,\exists n.\big(T(n)\land\forall(k < n).\neg T(k)\big)$
https://en.wikipedia.org/wiki/Induction,_bounding_and_least_number_principles
==== Logical definitions and theorems ====
$\forall(z < y).P(z)$ ... $\forall z.\big((z < y)\to P(z)\big)$
(used above)
$\forall x.\neg P(x)\leftrightarrow\neg\exists x.P(x)$
$(A\to\neg B)\leftrightarrow\neg(A\land B)$
$(A\to B)\to(\neg B\to\neg A)$
==== On Q ====
Theory signature/primitives: $S, 0$ and $+, \cdot$
https://en.wikipedia.org/wiki/Robinson_arithmetic
https://ncatlab.org/nlab/show/Robinson%20arithmetic
==== Axioms ====
(Universal closures $\forall x$ and, where applicable, $\forall y$)
$\neg(Sx=0)$
$(Sx=Sy)\to(x=y)$
$x+0=x$
$x+Sy=S(x+y)$
$x\cdot 0=0$
$x\cdot Sy=x\cdot y+x$
But one formula with an existential quantifier:
$y=0\lor\exists z.(Sz=y)$
==== Definitions ====
Here the relation is defined in terms of primitives that aren't relations.
$k < n$ ... $\exists m. k + Sm = n$
==== Some calculations ====
${\mathbb N}$ ... ${0, 1, 2, 3, \dots}$
$\frac{1}{2}0\cdot (0+1) = 0 = 0$
$\frac{1}{2}1\cdot (1+1) = 1 = 0 + 1$
$\frac{1}{2}2\cdot (2+1) = 3 = 0 + 1 + 2$
$\frac{1}{2}3\cdot (3+1) = 6 = 0 + 1 + 2 + 3$
Assume, for any particular $n$ ...
$\sum_{k=1}^n k = \frac{1}{2}n\cdot (n+1)$
then
$\sum_{k=1}^{n+1} k = \big(\sum_{k=1}^{n} k\big)+(n+1) = \frac{1}{2}n\cdot (n+1) + \frac{1}{2} 2 (n+1) = \frac{1}{2}(n+1)(n+2)$
==== Q Models ====
==== Standard model ====
The "intended" "standard" "matchbox" model of the naturals, call it $\mathbb N$.
==== One infinite number ====
The standard model plus one element, $\mathbb N+{c}$.
$x\cdot y=y\cdot x$
$x+y=y+x$
$c\cdot 0=0$
$c+x=c$
E.g. $c+S0=c$.
So $\exists y. c+Sy=c$.
I.e. $c < c$.
==== Certain polynomials ====
Integer-coefficient polynomials with positive leading coefficient, plus the zero polynomial.
(see Wikipedia)
==== On PA ====
https://en.wikipedia.org/wiki/Peano_axioms
$\forall n.\Big(Q(0)\land \forall n.\big(Q(n)\to Q(Sn)\big)\Big),\to,\forall m.Q(m)$
$\bullet$ Induction example:
$\forall(n\in{\mathbb N}). \sum_{k=1}^n k = \frac{1}{2}n\cdot (n+1)$
$\bullet$ Also proves that every number has a predecessor - see the axiom in ${\mathsf Q}$.
$\bullet$ Also e.g. proves that every number is either even or odd.
==== Complete induction ====
Complete induction:
$\forall n. \Big(\big(\forall(k < n).Q(k)\big),\to,Q(n)\Big),\to,\forall m.Q(m)$
Hint for proof from induction:
$\big(\forall(k < Sn).Q(k)\big),,\leftrightarrow,,\big(\forall(k < n).Q(k)\big)\land Q(n)$
==== Contrapositive ====
$\forall n. \Big(\big(\forall(k < n).Q(k)\big),\to,Q(n)\Big),\to,\forall m.Q(m)$
$\neg\forall m.Q(m),\to,\neg\forall n. \Big(\big(\forall(k < n).Q(k)\big),\to,Q(n)\Big)$
$Q$ ... $\neg T$
$\neg\forall m.\neg T(m),\to,\neg\forall n. \Big(\big(\forall(k < n).\neg T(k)\big),\to,\neg T(n)\Big)$
$\big(\neg\neg\exists m.T(m)\big),\to,\neg\neg\exists n.\Big(\big(\forall(k < n).\neg T(k)\big),\land T(n)\Big)$
==== a < a ====
For any relation $R$,
$P_R$ ... $\exists x.\big(x=a\land\forall y.(R(y,a)\to y\neq a)\big)$
$R(a,a)\to\neg P_R$
Least number principle for $R$ equal to $<$ and $T$ equal to $x=a$:
$\big(\neg\neg\exists m.m=a\big),\to,\neg\neg P_{ < }$
This exactly negates that an $a$ would have $a < a$.
==== Decidable T ====
Weakening the antecedant to $\exists m. T(m)$, we can scan the $k < m$ for a lowest element.
As always, adopt Markov's principle if ya like.
==== Non-standard models of PA ====
Note: Countable non-standard models
https://victoriagitman.github.io/images/PAmodel.jpg
(Where the division notation $/$ is division without rest.)
Roughly, ${\mathbb N}+{\mathbb Q}\times{\mathbb Z}$
Indeed we can take $\mathsf Q$ and add a recursive (computable) set of axioms,
saying that $c$ is bigger than each standard natural $n$.