Proof
(related to Proposition: Simple Consequences from the Definition of a Vector Space)
Let $V$ be a vector space over the field $F.$ Since, by definition, all vectors $x\in V$ form an Abelian group, the statements 1 to 8 follow immediately as a corollary from the group axioms. The statements 9 to 12 remain to be shown.
Ad 9)
 Since $1=1\cdot (1)$ in the field $F$, the axioms of scalar multiplication show that $1x=(1\cdot (1))x=1\cdot ((1)x)).$
 Since by these axioms, $1$ is the neutral element of the scalar multiplication in $V$, it follows that $1x=(1)x.$
Ad 10)
 Since $0=1+(1)$ in the field $F$, the axioms of scalar multiplication demonstrate that $0x=(1+(1))x=x+(x)=o,$ in which we have used the rules no. 2 and no. 9.
Ad 11)
 By no.2 we have $x+(x)=o$ for all vectors $x\in V.$
 Thus, by the axioms of scalar multiplication we have $\lambda o=\lambda(x+(x))=\lambda x + (\lambda x)=o$ for all $\lambda\in F.$
Ad 12)
 No. 10 and/or no. 11 imply $\lambda x=o.$
 The converse remains to be shown, so assume $\lambda x=o.$
 We want to show that then, at least one of $\lambda$ or $x$ must be zero.
 Assume $\lambda\neq 0.$ Then $\lambda^{1}\cdot\lambda=1.$ Thus, multiplying both sides of the above equation by $\lambda^{1}$ and applying the axioms of scalar multiplication as well as no. 11 we get $$\begin{array}{rcl}\lambda^{1}\cdot(\lambda x)&=&\lambda^{1} o\\(\lambda^{1}\cdot\lambda)x&=&o\\1x&=&o\\x&=&o.\end{array}$$
 Now, assume $x\neq o.$ Note that $\lambda x=o=\lambda (x+(x))=\lambda x+(\lambda x)=o+(\lambda x)=\lambda x.$
 Since $\lambda x=\lambda x$ and $x\neq o$, we must have $\lambda=\lambda$, which is only true for $\lambda=0.$
∎
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References
Bibliography
 Knauer Ulrich: "Diskrete Strukturen  kurz gefasst", Spektrum Akademischer Verlag, 2001
 Knabner, P; Barth, W.: "Lineare Algebra  Grundlagen und Anwendungen", Springer Spektrum, 2013