“If” versus “and” in statements

discrete mathematicslogicpredicate-logicpropositional-calculus

Let I(x) be the statement “x has
an Internet connection”and C(x, y) be the statement “x and y have chatted over the
Internet ”, where the universe of discourse for the variables x and y consists of all the
students in your class. Use quantifiers to express each of these statements.

  1. There is a student in your class who has chatted with everyone in your class over
    the Internet. Answer is :∃x∀y [x ≠ y → C(x, y)]
  2. There are at least two students in your class who have not chatted with the same
    person in your class. Answer is : ∃x∃y [x ≠ y ∧ ∀z ¬(C(x, z) ∧ C(y, z))]

I am confused about whether I should use "if" or "and". In the first question, "if" is used. In the second question "and" is used. But I would use "and" in the first statement.

Best Answer

Let's first do a different example:

Suppose I want to say that "All cubes are big"

Now, if I use:

$\forall x (Cube(x) \land Big(x))$

then I end up saying that everything is a cube and big: clearly not what I want: just because all cubes are big does not mean that everything is a cube. What I want is that if something is a cube, then it is big, i.e. I need to first make sure that I restrict myself to all cubes before I say anything about them:

$\forall x (Cube(x) \to Big(x))$

Likewise, in your problem, if you use:

$\exists x \forall y [x \neq y \land C(x, y)]$

you end up saying that for any $y$ it is true that $y \neq x$ and that $x$ has chatted with $y$: So now you end up saying that all students $y$ are different from $x$ ... which is obviously not the case, since $x$ is not different from $x$.

So again, you first need to restrict the $y$ to any student other than $x$ before you say that $x$ chatted with $y$. So that's why you use an $\to$: if $y$ is someone other than $x$, then $x$ will have chatted with $y$:

$\exists x \forall y [x \neq y \to C(x, y)]$

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