# How do you do First order models here?

So I have a question as follow:

Question 1

And I wonder if e.g. $$\exists x[R(x,x)]$$ means that there are things like , or does it mean that there are the same things (like twice in R) that are e.g. $$$$ and $$$$?

Also, my next question is if there is $$\exists x\exists y[R(x,y)]$$, does it mean there exist things in R like e.g. $$$$?

I would appreciate any help as I'm really confused about what it actually means and I can't find any resources online.

(I'll first note that there appears to be a couple of errors in this question - presumably the functions listed should be $$\delta_1$$ through $$\delta_4$$, rather than having two copies of $$\delta_2$$ and no $$\delta_4$$.)
$$\exists x [R(x, x)]$$ means "There is an element $$x$$ in our domain such that $$R(x, x)$$ holds, i.e. it is in a relation to itself.
Outside of this kind of problem, an example would be if we're working in real numbers and $$R(a, b)$$ means "$$b$$ is equal to the square of $$a$$". In this case, $$\exists x [R(x, x)]$$ is true, because we can take, for example, $$x = 1$$ and note that $$1^2 = 1$$, so $$R(1, 1)$$ is true.
For the ones with two qualifiers, always work from the outside in, reading as "[There exists / for all] $$x$$, such that [there exists / for all] $$y$$, R(x, y) holds". For example, $$\forall x \exists y [R(x, y)]$$ means "For every $$x$$ in the domain, there exists a $$y$$ such that $$R(x, y)$$ holds", i.e. "Every value in the domain is represented as the first member of at least one pair". By comparison, $$\exists x \forall y [R(x, y)]$$ means "there is an element $$x$$ in the domain such that $$R(x, y)$$ holds for every $$y$$ in the domain", i.e. there's a single element that pairs with every element in the domain.
Again going outside this problem, if $$R(x, y)$$ is "$$y$$ is the square of $$x$$" then $$\forall x \exists y [R(x, y)]$$ is true because given any real value $$x$$ we can find a $$y$$ such that $$y = x^2$$, but $$\exists x \forall y [R(x, y)]$$ is definitely false because there isn't one single value of $$x$$ such that every $$y$$ is its square. On the other hand, if $$R(x, y)$$ represented "$$x$$ is less than the square of $$y$$", then $$\exists x \forall y [R(x, y)]$$ is true because we can pick, for example, $$x = -1$$ and it is true that it is less than the square of every real number.