I don't get why people are pointing ∃x∃y(R(x,y)Λ~ R(y,x)) to be Anti-symmetric.
Okay if I follow above notion of anti-Symmetry
Consider a set A={1,2,3,4}
And relation S={(1,2) , (2,3) , (3,2) }
The Matrix of this relation is
$\begin{pmatrix} 0&1 & 0&0 \\ 0 & 0& 1&0 \\ 0 & 1 & 0 &0\\ 0 & 0 &0 &0 \end{pmatrix}$
This relation is neither symmetric nor Anti-symmetric.
Because for symmetry, the off-diagonal 1's have a 1 in corresponding mirror-image treating diagonal as the mirror.
So, for (1,2) we don't have (2,1)
And, for Anti-Symmetry, we allow diagonal pairs (because of x=y) but only take either lower off-diagonal elements or upper off-diagonal elements but not both at a time for Anti-Symmetry.
But, according to definition of ∃x∃y(R(x,y)Λ~ R(y,x))
we see (1,2) satisfies this and hence it marks this relation as anti-symmetric. Which is false.
For anti-Symmetry, we have ∀x∀y(R(x,y)↔R(y,x))→(x=y)
For Asymmetry, we have ∀x∀y(R(x,y)→~R(y,x))
I think to go by as what Arjun sir said.
∀x∀y(R(x,y)⟹R(y,x))
This implies symmetry but since this does not hold true for all relations, so it is satisfiable but not valid.
It's negation ∃x∃y(R(x,y)Λ~ R(y,x)) also does not hold true for all relations, but for the relation shown as example above, this negation is satisfied. Hence, satisfiable but not valid.
So, answer option (B)