Is there any false case for that: $\exists x \in D, \forall y \in D, P(x, y) \implies P(y, x)$?

Question

Is there any false case for that: $\exists x \in D, \forall y \in D, P(x, y) \implies P(y, x)$?\

I just can get the true case.\

How can we define D and P in a false case?\

Thx guys.

Answer 1

Rock, paper, scissors? Let $P(x,y)=x$ beats $y$. Rock beats scissors, but scissors doesn't beat rock. Scissors beats paper but paper doesn't beat scissors. Paper beats rock, but rock doesn't beat paper.

Answer 2

$D = \varnothing$ seems like a pretty obvious (vacuous) falsification.

Answer 3

I assume that in the formula : $∃x∈D,∀y∈D,P(x,y) \rightarrow P(y,x)$ (call it : (A)) the scope of the quantifiers is the conditional, and not only the antecedent; i.e. :

(A) --- $∃x∀y (P(x,y) \rightarrow P(y,x))$.

If so, assume as domain $D$ of the interpretation the set $\mathbb N$ of natural numbers and interpret the predicate $P$ as the relation "less than" ($<$).

Consider the negation of the formula above, i.e. :

$\lnot ∃x∀y (P(x,y) \rightarrow P(y,x))$,

which is :

$∀x∃y \lnot (P(x,y) \rightarrow P(y,x))$.

By the equivalence of $A \rightarrow B$ with $\lnot A \lor B$ and double negation, this is equivalent to :

$∀x∃y (P(x,y) \land \lnot P(y,x))$.

We have that :

$(n < n+1) \land \lnot (n+1 < n)$

is true for every $n \in \mathbb N$.

Thus :

$\exists y ((n < y) \land \lnot (y < n))$

is true for every $n$, i.e.

$\forall x\exists y ((x < y) \land \lnot (y < x))$

is true in $\mathbb N$.

In conclusion, we have found an interpretation in which the negation of the formula (A) above is true; thus, in this interpretation, (A) is false.