Is there any false case for that: $\exists x \in D, \forall y \in D, P(x, y) \implies P(y, x)$? 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.
 A: 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.
A: $D = \varnothing$ seems like a pretty obvious (vacuous) falsification. 
A: 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. 
