The intersection of difference sets complement an array of union

ES5 wording

There are a = [1,2,3,4,5]

var b = [2,4,6,8,10]

// // intersection

var c = a.filter(function(v){ return b.indexOf(v) > -1 })

// // difference set

var d = a.filter(function(v){ return b.indexOf(v) == -1 })

// //Complement

var e = a.filter(function(v){ return !(b.indexOf(v) > -1) })

 .concat(b.filter(function(v){ return !(a.indexOf(v) > -1)}))

// union

var f = a.concat(b.filter(function(v){ return !(a.indexOf(v) > -1)}));

console.log ( "Array a:", a);

console.log ( "Array b:", b);

console.log ( "a and the intersection b:", c);

console.log ( "a difference between the set and b:", d);

console.log ( "a and b complement:", e);

console.log ( "a to b and set:", f);

 

With the wording ES6

A = var [1,2,3,4,5] 
var B = [2,4,6,8,10] 
the console.log ( "Array A:", A); 
the console.log ( "Array b:" , B); 
var new new SA = the set (A); 
var the set new new SB = (B); 
// intersection of 
the let INTERSECT = a.filter (X => sb.has (X)); 
// set difference 
let minus = a.filter (X => sb.has (X)!); 
// complement 
let complement = [... a.filter (! x => sb.has (x)), ... b.filter ( = X> sa.has (X))];! 
// set and 
the let unionSet = Array.from (the set new new ([... a, ... b])); 
the console.log ( "a and b intersection: ", INTERSECT); 
the console.log (" a difference between the set and b: ", minus); 
the console.log (" complement of a and b: ", complement); 
the console.log (" a and b union: ", unionSet);