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$\textbf{Practice 3.2.3}$ $\textbf{Theorem}$ $xy=0$ or $y=0$ iff $x=0$ or $y=0$ $\textbf{Proof}$ If $x=0$ or $y=0$, then $xy=0$ by $3.2.2(b)$ and $M2$ Conversely suppose that $xy=0$ By tautology $(1.3.12 (p)$ it suffices to show that $y=0$. Since $y=0$ Since $x \neq 0$ $1/x$ exists by $(a)$ $M4$ Thus $$ 0 = \frac{1}{x} \cdot 0 \, \, \text{by} \, A5$$ $$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 0 = \frac{1}{x} \cdot xy \, \, \text{since} \, xy=0 $$ $$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 0 = \Big( \frac{1}{x} \cdot x \Big) y \, \, \text{by (c)} \text{DL}$$ $$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 0 = \Big( x \cdot \frac{1}{x} \Big)y \text{by} \, \, \, \, \, \, \, \, (d) \, \, \, \text{M5}$$ $$\, \, \, \, \, \, \, \, 0 = 1 \cdot y \, \, \, \text{by (f) A5} $$
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Thu, 16 Sep 2021 02:46 GMT