Would this Interpretation prove this argument to be deductively invalid (FOL)

Question

I want to show:

$$∀x(Ax → ∃y(By ∧ Rxy)) ∴ ∀y(By → ∃x(Ax ∧ Rxy))$$

Is deductively invalid. Would the following interpretation show this?

Domain: all natural numbers

Let $Ax$ be $x=2$

Let $By$ be $y=0$

Let $Rxy$ be $x$ is less than $y$

Under this interpretation, would the premise be true and the conclusion false?

Answer 1

I think that interpretation is not a good counterexample because, as you said, the premise needs to be true and the conclusion false to show that the argument is invalid. However, in your interpretation, the premise itself starts out false, because it says that:

∀x(x=2 → ∃y(y=0 ∧ x<y)) (in words: For any natural number equal to 2, there is at least one natural number equal to 0 that is greater than it), and if you apply the rule of Universal Instantiation (the rule that lets you drop the Universal Quatifier), you can get: 2=2 → ∃y(y=0 ∧ 2<y), and since the antecedent of this conditional is true, by modus ponens, you get the consequent: ∃y(y=0 ∧ 2<y) (there is at least one natural number equal to 0 and greater than 2), which is false, so it doesn't matter if the conclusion is true or not, the interpretation will not be a counterexample that shows that the argument is invalid.

To show the argument is invalid, I suggest this interpretation:

Ax="x is a famous soccer player", By="y is a televised game", Rxy="x has participated in y"

With that you get the premise: ∀x(x is a famous soccer player→∃y(y is a televised game ∧ x has participated in y)), or more naturally: all famous soccer players have participated in a televised game. And the conclusion would be: ∀y(y is a televised game→∃x(x is a famous soccer player ∧ x has participated in y)), or more naturally: In every televised game, there is at least one famous soccer player that participates in it, which is clearly false in the case of televised tennis games, basketball games, etc. but maybe someone else with a better imagination than mine might come up later with a better counterexample.

Answer 2

No it is no counterexample. Both the premise and the conclusion is wrong for this model.

I put $$∀x(Ax→∃y(By∧Rxy))∴∀y(By→∃x(Ax∧Rxy))$$ in the TreeProofGenerator and it tells me that the statement is invalid and shows the following

Countermodel:
Domain: { 0 }
A:  { }
B:  { 0 }
R:  { (0,0) }

that actually shows that the statement is wrong.