Anti-symmetric or asymmetric for a relation between pairs in the set of Z x Z?

Question

Is this anti-symmetric or asymmetric?

I at first thought asymmetric because anti-symmetric would mean a = c and b = d which would not be true.

But because the domain is the Cartesian product of all integers I am not sure we can say that a = c and b = d.

Answer 1

Symmetric: $$\forall \langle a,b\rangle \in \Bbb Z^2\; \forall \langle c,d\rangle \in\Bbb Z^2 \;\Big(\langle a,b\rangle \mathsf R\langle c,d\rangle \implies \langle c,d\rangle \mathsf R\langle a,b\rangle \Big)$$

So is it so?: $$\forall \langle a,b\rangle \in \Bbb Z^2\; \forall \langle c,d\rangle \in\Bbb Z^2 \;\Big((a<c\wedge b<d) \implies (c<a\wedge d<b)\Big)$$

---

Antisymmetric: $$\forall \langle a,b\rangle \in \Bbb Z^2\; \forall \langle c,d\rangle \in\Bbb Z^2 \;\Big(\big(\langle a,b\rangle \mathsf R\langle c,d\rangle \wedge\langle c,d\rangle \mathsf R\langle a,b\rangle \big) \implies \langle a,b\rangle =\langle c,d\rangle \Big)$$

So is it so?: $$\forall \langle a,b\rangle \in \Bbb Z^2\; \forall \langle c,d\rangle \in\Bbb Z^2 \;\Big(\big(( a<c \wedge b<d )\wedge( c<a \wedge d<b ) \big)\implies \big( a=c \wedge b=d \big)\Big)$$