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The statement i want to translate is this: x is the smallest real number and P(x) is false

$\exists x \in \mathbb{R} \forall y \in \mathbb{R} \neg (P(x)); x > y$

I don't know how to put two statements in one predicate sentence.

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+1 for showing your work and getting the "hang" of formatting! – amWhy Oct 30 '12 at 22:19
up vote 2 down vote accepted

You don’t want the existential quantifier: ‘$x$ is the smallest real number’ is simply $$\forall y\in\Bbb R\Big(\lnot(y<x)\Big)\;.$$ Since ‘$P(x)$ is false’ is $\lnot P(x)$, the conjunction of the two is simply

$$\forall y\in\Bbb R\Big(\lnot(y<x)\Big)\land\lnot P(x)\;.$$

This says that some $x$ that was presumably specified previously has the desired properties.

The existential quantifier is needed if you want to say that such an $x$ exists:

$$\exists x\left(\forall y\in\Bbb R\Big(\lnot(y<x)\Big)\land\lnot P(x)\right)\;.$$

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Note that you can also write the equivalent (to middle statement) by stating, essentially, there does not exist a y such that $y < x$ and not $P(x)$, i.e. $ \lnot \exists y \in \mathbb{R} \left(y < x\right) \land \lnot P(x)$. – amWhy Oct 30 '12 at 22:12
"$x$ is the smallest $R$" implies uniqueness, $(\forall y \in R)x \leq y$ doesn't. So the first translation won't quite do, strictly speaking. – Peter Smith Oct 30 '12 at 22:22
Also, if you want to say that a real number x exists such that the above is true, then simply enclose the above in parentheses and put "$\exists x \in \mathbb{R}$" in front of $(\lnot \exists y \in \mathbb{R} \left(y < x\right) \land \lnot P(x))$. – amWhy Oct 30 '12 at 22:24
@Peter: That depends on what is assumed about $\Bbb R$. I took it as given that $<$ is a linear order on $\Bbb R$. If one does not make some assumption, one ends up writing down axioms for $\langle\Bbb R,<\rangle$ as (at least) a linear order. – Brian M. Scott Oct 30 '12 at 22:25
@Brian I'd say, qua logician, that there is a difference between translating (what the OP was after) and giving a sentence that provably equivalent on non-logical background assumptions. The canonical translation of something of the form "$a$ is the $F$" using standard first-order apparatus should run $Fa \land \forall y(Fy \to y = a)$ or the like. – Peter Smith Oct 31 '12 at 13:11

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