Shortest function that gives same result as === when comparing 2 variables

Question

This is my take on it. Is it possible to make it shorter? (Not just by turning it into an arrow function) Must not use any ===

this is my third attempt, I really don't know now if its possible to shorten it evem further

function strictEquals(a, b) {
  if ([a, b].every(isNaN)) return false;
  return Object.is(a + 0, b + 0) || Object.is(a, b);
}

second attempt.

function strictEquals(a, b) {
  if (Number.isNaN(a) && Number.isNaN(b)) {
    return false;
  }
  if (
    (Object.is(a, 0) && Object.is(b, -0)) ||
    (Object.is(a, -0) && Object.is(b, 0))
  ) {
    return true;
  }
  return Object.is(a, b);
}

you can test yours with this battery of tests

console.log('(1, 1), expected true', strictEquals(1, 1));
console.log('(NaN, NaN), expected false', strictEquals(NaN, NaN));
console.log('(0, -0), expected true', strictEquals(0, -0));
console.log('(-0, 0), expected true', strictEquals(-0, 0));
console.log("(1, '1'),expected false", strictEquals(1, '1'));
console.log('(true, false), expected false',strictEquals(true,false));
console.log('(false, false), expected true', strictEquals(false,false));
console.log("('Water', 'Oil'), expected false", strictEquals('Water', 'Oil'));
console.log('(Undefined, NaN), expected false', strictEquals(undefined, NaN));

this one was a mistake (my first attempt)

function strictEquals(a, b) { if ((Number.isNaN(a) && Number.isNaN(b)) || !Object.is(a, b)) return false; return Object.is(a + 1, b + 1); }

Answer 1

Modified version:

function strictEqual(a,b) {
  return (Object.is(a, b) || Object.is(a+1, b+1)) && !Number.isNaN(a)
}