4 divides 8 but 8 does not divide 4 implies it not symmetric. Here we don't have to check for anti-symmetry property because the left side of implication is false because as per definition of anti-symmetry :

if aRc and cRa, then a = c. It is represented as :

aRc ∧ cRa → a = c. We know that if left side of implication symbol is false, then the implication statement is always true.

So, the statement becomes :

(4 divides 8 and 8 divides 4 ) → ( 4 = 8 ).

Here 4 divides 8 and 8 divides 4 . This statement is false as 8 does not divide 4 over set of integers and so left side of implication is false, and hence the statement is true. So, it is anti-symmetric…...

Now comes for the sets.

The one Shaik explained i.e. {1,2} R { 1,2}, follows reflexive so this element of relation is also anti-symmetric.

Now coming to my example :

Courtesy : Kenneth Rosen Discrete Maths.

Way before the multiset chapter was introduced in Kenneth Rosen Discrete Maths book, the above lines were said just at the start of the set chapter regarding equality of sets, i.e. the order and number of times an element occur does not matter.

So, as per definition of anti-symmetry :

if ( {1,3,5} is related to { 1,3,3,3,5,5,5,5} ∧ { 1,3,3,3,5,5,5,5} is related to {1,3,5} ) → ( {1,3,5} is equal to { 1,3,3,3,5,5,5,5} ).

U can see that both sides of the implication are true and as per logic, True → True is always True.

Hence, the condition of anti-symmtry gets satisfied…..