POSET always satisfy the property of reflexivity , anti-symmetricity and transitivity . So all these properties can be used for checking S1 and S2 .
Here in S1 pairs (2,3) and (3,2) are violating the anti-symmetric property hence S1 not a POSET.
In S2 there is no such violating pair and all of the above three properties satisfy , hence S2 is POSET.