Which of first or second input are we comparing the third to? If both, then I don't think you operator makes sense, since that would suggest your ~=~ operator is transitive. Since all inputs are comparable, I assume, that would mean your operator is an equivalence relation.
Due to transitivity:
A ~=~ B ~=~ C and
C ~=~ E ~=~ F implies that
A ~=~ F which would not necessarily be the case.
For instance:
Let X be a large set with many elements. Then a "reasonable intelligent human" might suggest that:
, even if this is actually the case, you expect it not to be for some large enough number. Then your operator cannot be transitive, but then what use is the knowledge that input #1 is approximate equal to input #2?
But since you already "discovered" the algorithm, why not just show it, or the pseudo code, such that we could discover its true name?
Due to transitivity:
A ~=~ B ~=~ C and
C ~=~ E ~=~ F implies that
A ~=~ F which would not necessarily be the case.
For instance:
Let X be a large set with many elements. Then a "reasonable intelligent human" might suggest that:
(X `union` {1}) ~=~ (X `union` {2})
And also:
(X `union` {2}) ~=~ (X `union` {3})
...... (X `union` {999999999999}) ~=~ (X `union` {1000000000000 })
which implies
(X `union` {1}) ~=~ (X `union` {1000000000000 })
, even if this is actually the case, you expect it not to be for some large enough number. Then your operator cannot be transitive, but then what use is the knowledge that input #1 is approximate equal to input #2?
But since you already "discovered" the algorithm, why not just show it, or the pseudo code, such that we could discover its true name?