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I'd like to ask if someone can help me out with proof of this statement. I have to proof this statement for x,y,z $\in$ $\Bbb N$

I was thinking about proving the (∃z) part. By showing atleast 1 situation where it is true.

(∀x)(∀y)(∃z)((z|x)∧(z|y)∧((∀u)((u|x ∧ u|y) -> u|z)))

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up vote 3 down vote accepted

If I've understood your question (I added a few brackets to help disambiguate what is implying what), you need to prove (or disprove, using a counterexample):

For $x, y, z \in \mathbb{N}$

$$(\forall x)(\forall y)(\exists z)\{[(z|x) \land (z|y) \land (\forall u)((u\in \mathbb{N}) \land u|x \land u|y)] \rightarrow (u|z)\} $$

In words:

For all $x, y \in \mathbb{N}$, there exists a $z\in \mathbb{N}$ such that,

Premise: IF ($z$ divides $x$ and $z$ divides $y$, such that for all $u\in \mathbb{N}$, if $u$ divides $x$ and $u$ divides $y$),

Consequent: (all such) $u$ divides (such a) $z$.


Here's where the definition of divisibility of numbers in $\mathbb{N}$ comes in. You need to "unpack" what the premise is stating, in terms of the definition of what it means to assert $a|b$ for $a, b \in \mathbb{N}$, and how that the premise thereby implies the conclusion (consequent).

Start with the premise: Assuming the premise, try to construct an argument "in words" first, using your definition, and proceeding step by step, deriving the conclusion. You need to prove that given all $x, y \in \mathbb{N}$, if there exists at least one $z\in \mathbb{N}$ such that $z$ divides $x$ and $z$ divides $y$, and if any $u \in \mathbb{N}$ also divides $x$ and divides $y$, then it must be the case that $u$ divides $z$.

Then translate that argument into logical statements that follow from the premise, and imply the conclusion. If you can derive the conclusion from the premise, then you will have proven the conclusion.

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