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Universe the real numbers.

  1. between any integer and any larger integer there is a real number.

$\forall x \forall y$ $(x \in Z \wedge y \in Z \wedge y>x \rightarrow \exists k (x<k<y)$

  1. Every nonzero real number is positive or negative.

$\forall x(x\not=0 \rightarrow x>0 \lor x<0$

3. All people hate some taxes.

$\forall x \exists y$( if x is a person and y is a tax then x hates y)

Fernando Martinez
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1 Answers1

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You need to fix (add) parentheses, and on number $(1)$, you need to say that $k \in \mathbb R$:

$$\forall x\forall y\Big((x \in \mathbb Z \wedge y \in \mathbb Z \wedge y>x )\rightarrow \exists k (k \in \mathbb R \land x< k <y)\Big)$$

Same thing for $(2)$. Here we need to specify $x \in \mathbb R$:

$$\forall x\Big((x\in \mathbb R \land x\not=0) \rightarrow (x>0 \lor x<0)\Big) $$ For $(3)$, $$\forall x\Big(\text{Person}(x)\rightarrow \exists y(\text{Tax}(y) \land \operatorname{Hates}(x, y))\Big)$$

amWhy
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  • I see so you cannot have in the antecedent both x is a person and y is a tax because In english what is being said is " All people hate some taxes" "if someone is a person, then there exists a tax y such that this person hates the tax" – Fernando Martinez Jun 12 '14 at 16:47
  • Exactly, Fernando. – amWhy Jun 12 '14 at 16:47
  • This translation can be a bit confusing.... – Fernando Martinez Jun 12 '14 at 16:48
  • So if you had "All people hate all taxes" It would be $\forall$ x( x is person $\rightarrow$ $\forall y$ y is a tax x hates y – Fernando Martinez Jun 12 '14 at 16:51
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    It becomes easier to understand, with practice, and it helps if you put down your answers, for a day or so, and then return to them to read back exactly what you wrote, and see whether it states the original sentence. You may spot some missing info, or incorrect translations, that way. – amWhy Jun 12 '14 at 16:54
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    In that case, the universally quantified $y$ can be placed at the start to get $$\forall x \forall y([P(x) \land T(y)]\rightarrow H(x, y) \equiv \forall x (P(x)\rightarrow \forall y (T(y) \rightarrow H(x, y)))$$ – amWhy Jun 12 '14 at 16:57