Given $R = \{ ( 1,2 ) \}$ over $A = \{ 1,2,3,4 \}$:
$R$ is not reflexive because there's no $(1,1)$ and $(2,2)$.
$R$ is not symmetric because there isn't $( 2,1 )$.
$R$ is anti-symmetric because there isn't $( 2,1 )$.
And $R$ is transitive because there isn't a $(2,x)$ element in $R$.
If $(1,2) \in R \land (2,x) \in R$ is false, then
$(1,2) \in R \land (2,x) \in R \to (1,x) \in R$ is true,
and therefore $R$ is transitive.
Is this correct?
The relation is not reflexive, since $(a,a) \not \in R$ for every $a \in \{1,2,3,4\}$.
The relation is not symmetric, since while $(1,2) \in R$, the reversal $(2,1) \not \in R$.
The relation is anti-symmetric in a vacuous sense: since $(1,2) \in R$ and is the only ordered pair in $R$, there is no "violation" in the sense that since there is no pair of related pairs $(a,b),(b,a) \in R$ with $a \ne b$ to begin with. Thus, anti-symmetry holds trivially.
The relation is transitive, for similar reasons. If $R$ were not transitive, then there would be pairs $(2,x),(x,y) \in R$ for which $(1,y) \not \in R$. However, no $(2,x) \in R$ for any $x$, so there is no issue there. Vacuously, in a sense, $R$ thus is transitive.
...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.
Not reflexive since $(1,1,),\ldots,(4,4)$ are missing, and not symmetric since $(2,1)$ is missing, but both antisymmetric and transitive since premises of implications (in their definitions) are not fulfilled.
