Matematické Fórum

jozou · 05. 09. 2011 18:41

Zdravim,

mam jeden problem s axiomatizaciou realnych cisel. Popis je v anglictine
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Suppose following axiomatization of real numbers:
( $\exists !$ is syntactical sugar for uniqueness quantifier)

(+ 1) $\forall a \forall b~a+b = b+a$
(+ 2) $\forall a \forall b\forall c~ a+(b+c) = (a+b)+c$
(+ 3) $\forall a \forall b \exists ! x~a+x = b$

(. 1) $\forall a \forall b~ab = ba$
(. 2) $\forall a \forall b \forall c~a(bc) = (ab)c$
(. 3) $\forall a \forall b~(a \neq 0 \rightarrow \exists ! x~ax = b)$

(+ .) $\forall a \forall b \forall c~(a+b)c = ac + bc$

(< 1) forall $a$ forall $b$ , one of the following holds: $a < b$ , $a = b$ , $b < a$
(< 2) $\forall a \forall b \forall c~a<b \wedge b < c \rightarrow a < c$
(+ <) $\forall a \forall b \forall c~a<b \rightarrow a+c < b+c$
(. <) $\forall a \forall b~0 < a \wedge 0 < b \rightarrow 0 <ab$

(sup) Let $A$ be non-empty set, s.t. it has upper bound. Then $\exists \alpha (\forall x~x \leq \alpha~\wedge~\forall \beta ~ \beta < \alpha \rightarrow ~ \exists a~ \beta < a)$

Now following theorem holds: There is precisely one element (which will be denoted $0$ ), which is solution of equation $\forall a~ a+x = a$ .
Proof: Assume some fixed $b$ . Let the only solution of $b+x = b$ be denoted by symbol $0$ . So $b+0=b$ holds.

Now proof is straightforward and I will not finish it. My problem is that $0$ is formal symbol of our first-order theory (i.e. it is non-logical constant) and no axiom define any property that should $0$ obbey. However the first step of the theorem says something like $\exists b~b+0 =b$ . My question is why is this correct? I don't see any inference rule which interlinks $0$ and axiom (+ 3).

any ideas why is that correct? thanks