iv. triangle inequality , 
\(d\left(x,y\right)\le d\left(x,z\right)+d\left(z,y\right)\) 
Proof by proving that counter example is not possible
Let us assume that JD(x,y)>JD(x,z)+JD(z,y)
\(\Rightarrow1-\frac{\left|x\cap y\right|}{\left|x\cup y\right|}>1-\frac{\left|x\cap z\right|}{\left|x\cup z\right|}+1-\frac{\left|z\cap y\right|}{\left|z\cup y\right|}\)
For this to be true \(x\cap y\) has to be non zero i.e it is not an empty set.Since JD(x,y) does not depend on z, thus removing all elements from z would only increase the value of JD(x,z)+JD(z,y) because it decreases the size of the denominator in both the terms therefore not affecting the inequality.Also, we can  remove all the element from z that are in x or y but not in \(x\cap y\)  as this operation would have no effect on inequality because it decreases the value of  \(\frac{\left|x\cap z\right|}{\left|x\cup z\right|}+\frac{\left|z\cap y\right|}{\left|z\cup y\right|}\) . .After the operations we can say that z is a subset of \(x\cup y\)
   Here z is the counterexample to the claim of metric distance but clearly it is not possible because it lies completely in \(x\cap y\).
Therefore we can say that there can be no z for which  \(d\left(x,y\right)\le d\left(x,z\right)+d\left(z,y\right)\)  .
Since JD(X,Y) satisfies all the 4 properties for being a metric , it is a metric.