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Considering the following field axiom:

$\forall$$x((x\neq0) \rightarrow \exists y(x.y=1))$

I am trying to show that this axiom is true for the model R = ($\mathbb{R}, 0, 1, +, .)$

Using some rules of satisfiability. Let (M, v) be a generic L-Model, then:

$(M, v) \vDash \forall$$x((x\neq0) \rightarrow \exists y(x.y=1))$ iff

$(M, [v]^{a}_x) \vDash $ $(x\neq0) \rightarrow \exists y(x.y=1)$, for all a $\in \mathbb{R} $

So:

1 - $(M, [v]^{a}_x) \vDash $ $¬(x\neq0) $, for all a $\in \mathbb{R} $

or

2 - $(M, [v]^{a}_x) \vDash $ $ \exists y(x.y=1)$, for all a $\in \mathbb{R} $

Considering 1, simplifying the expression:

$(M, [v]^{a}_x) \vDash $ $(x=0) $, for all a $\in \mathbb{R} $. So $a = 0$ for all a $\in \mathbb{R} $ is false.

Considering 2, we have:

$(M, [v]^{a}_x) \vDash $ $ \exists y(x.y=1)$, for all a $\in \mathbb{R} $

$(M, [v]^{a b}_{x y}) \vDash $ $(x.y=1)$, for all a $\in \mathbb{R} $, for some b $\in \mathbb{R} $. Finally:

$a.b=1$, for all a $\in \mathbb{R} $, for some b $\in \mathbb{R} $. This is not true when $a=0$.

So neither 1 or 2 are true following the above solution. What am I doing wrong in this case?

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    $\begingroup$ In English, use field (as in "campo") rather than body (as in "cuerpo"). $\endgroup$ – Andrés E. Caicedo Aug 20 '16 at 14:17
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You want to show that for every $a\in \mathbb{R}$ either 1 or 2 true is the case. You are correctly claiming that 2 is not the case only if $a=0$, but it is preceisely the case where 1 is true, so actually have shown what you wanted.In more details: if $a=0$ then really $(M,[v]^a_x)\nvDash \exists y (x*y=1)$ however 1 is true, meaning that $(M,[v]^a_x)\models x=0$, which is enough to conclude $(M,[v]^a_x)\models x\neq 0\to \exists y(y*x=1)$.

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  • $\begingroup$ What a mistake I made! Thanks for the answer! $\endgroup$ – Bruno A Aug 20 '16 at 13:53

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