Question concerning predicate Logic

Question

how do I say in predicate logic: $\forall x > 2 $ there exists ...

Thanks for every help!

Answer 1

Given $P(x, y)$ is a predicate over $\mathbb{U} \times \mathbb{V}$, we have $$(\forall x \in \mathbb{U})(\exists y \in \mathbb{V})(P(x, y)) $$

For this question, you would have something like $$(\forall x > 2)(\exists y \in \mathbb{R})(P(x, y)) $$ Assuming $\mathbb{R} \times \mathbb{R}$

We can read it as for all $x>2$ there exists some $y$ that $P(x, y)$ is a predicate, therefore, $(\forall x > 2)(\exists y \in \mathbb{R})(P(x, y)) $ is a preposition.

Answer 2

If $P$ is some property and I want to write "For all $x>2$ there exists $y$ satisfying $P$" formally in first-order predicate logic, then I can write $\forall x(x>2\rightarrow \exists yP)$. There are other abbreviations for this that are usually well understood, such as $(\forall x>2) (\exists y) P$, etc.

But I wanted to add a supplementary answer based on the discussion of $\rightarrow$ vs $\Rightarrow$.

I can't say for sure, but your confusion about these two symbols might be due to the habit some people have in writing proofs, where they write $A\Rightarrow B$ to denote "I know $A$, and now I'm concluding $B$." For example I might give my students the following exercise.

Suppose $x$ and $y$ are odd integers. Prove that $x+y$ is even.

Now, due to a mixture of laziness and over-excitement with using math symbols for the first time, many students will write a proof like this:

$x$ and $y$ are odd integers
$\Rightarrow$ $x=2a+1$ and $y=2b+1$ for some integers $a,b$
$\Rightarrow$ $x+y=2(a+b+1)$
$\Rightarrow$ $x+y$ is even

So, in each line, $A\Rightarrow B$ is signifying "$A$ is true. And from this I'm now concluding $B$."

Moreover, sometimes profs will use something like this notation when lecturing on the board (usually to save time writing out too many words).

But, in any case, this use of $\Rightarrow$ is only a convention. By itself, $A\Rightarrow B$ means "$A$ implies $B$", which is to say "If $A$ is true then $B$ is true." One should not assume that $\Rightarrow$ is signifying anything more than just this, without double-checking the surrounding context.