a←  1      + 0j1×1+⎕ct
   b← (1+⎕ct) + 0j1×1+⎕ct×2

   a = b        ⍝ therefore  (|a-b)     ≤ ⎕ct×(|a)⌈|b
1
   (|a) = |b    ⍝ therefore  (|(|a)-|b) > ⎕ct×(|a)⌈|b
0
                ⍝ therefore  (|(|a)-|b) > |a-b

But that contradicts the triangle inequality!