Confusion about proof of every asymmetric relation being an irreflexive relation

Question

Let R be asymmetric. We need to show R is irreflexive.

So by definition, we assume:

1. (x,y)∈R⟹(y,x)∉R

Definition of irreflexivity:

2. (x,x)∉R

Let's do an indirect proof, so we assume:

3. (x,x)∈R

Use Modus Ponens on 1 and 3, we get (x,x)∉R - contradiction. QED

The modus ponens step is where I get confused: the antecedent of the conditional is (x,y)∈R, but we are replacing it with (x,x)∈R instead. If the variables are different, how could we do that?

I suspect I may have confusion about the fundamental meaning of a variable, because by intuition this seems like a legitimate step, but I want to make sure I get it 100%. Could anyone help please?

Answer 1

As you correctly point out, your problem comes from not understanding variables properly. It is invalid to use $x,y$ without stating what it is.

Given any asymmetric relation $R$ on a collection $S$:

  For any $x,y \in S$:

    $(x,y) \in R \rightarrow (y,x) \notin R$.

  For any $x \in S$:

    $(x,x) \in R \rightarrow (x,x) \notin R$.

    If $(x,x) \in R$:

      $(x,x) \notin R$.

      Contradiction.

    Therefore $(x,x) \notin R$.

  Therefore $R$ is irreflexive.

Answer 2

x and y are arbitrary variables. Just because they are denoted by different letters does not mean that they must have different values.

Answer 3

When you say "For all x and for all y..." you are NOT assuming that x and y are not names for the same thing.

Answer 4

To expand on the others answers. This is your property :

$$\forall (x,y)\in E^2, ( (x,y)\in R \Rightarrow (y,x) \not\in R )$$

So this imply that

$$\forall t, ( (t,t)\in R \Rightarrow (t,t) \not\in R )$$

But you have that

$$\forall t \in E, ( (t,t) \not\in R \vee (t,t)\in R) $$

Hence

$$\forall t \in E, ( (t,t) \not\in R \vee ( (t,t) \in R \Rightarrow (t,t)\not\in R) \vee (t,t)\in R ) $$

So by modus ponens, you get

$$\forall t \in E, ( (t,t) \not\in R \vee (t,t)\not\in R) $$

And this is

$$\forall t \in E , (t,t) \not\in R $$