Is $\forall x \forall y (x \neq y \rightarrow (P(x,y) \leftrightarrow \neg P(y,x)))$ valid?

Question

I often came forward questions like these ( true or false) in my past papers. They are supposed to be answered quickly , not by calculations by "trying to make sense by intuition if they provide an obvious fact ( so they are true) or if we suspect that they're not, we may come up easily with a counterexample. But since there no official solutions outhere, I can't rely on me , to be sure whether I assumed something right . So here is my approach:

• If it is valid , then it's negation must be always false

• $\rightarrow \neg[\forall x \forall y (x \neq y \rightarrow (P(x,y) \leftrightarrow \neg P(y,x))) ] \equiv \exists x \exists y(x \neq y \land \neg(P(x,y) \leftrightarrow \neg P(y,x)) $

• In order this to be always false :

• assuming ** $x \neq y$ is always false , so we are in a universe with only one element , but this is not neccessary hence we move to the next case (2) (there are many cases that the statement is true)

• $\neg(P(x,y) \leftrightarrow \neg P(y,x))$ has to be always false $\rightarrow (P(x,y) \leftrightarrow \neg P(y,x))$ has to be always true which I don't see why if for example P(x,y) is defined as : "x is connected to y in a digraph" in a graph

So I came to conculsion, that there are cases in order the negation of the statement to hold true . Hence , it is not valid. Let me know if you agree or if you have another way of thinking of it

Answer 1

Suppose the domain of $x, y$ is the set of all human beings. Let $P(x, y)$ mean $x$ is married to $y$. Then $$\forall x \forall y (x \neq y \rightarrow (P(x,y) \leftrightarrow \neg P(y,x)))\,\text{ is not valid}.$$

Validity does depend on $P(x, y)$. For example, suppose the domain of $x, y$ is all integers. And define $P(x, y)$ as $x \gt y$. Then $$\forall x \forall y (x \neq y \rightarrow (P(x,y) \leftrightarrow \neg P(y,x)))\, \text{ is indeed valid}.$$

But with no pre-specified definition of $P(x, y)$, the statement has a counterexample, and cannot be considered valid.

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Note the first relation $P(x, y)$ is a symmetric relation, and is not anti-symmetric. The second relation $P(x, y)$ is assymmetric, and trivially, antisymmetric.