User blog comment:Wythagoras/Dollars Function/@comment-11227630-20130928111712/@comment-10429372-20130928143820

1. A group of zeroes

2. 2$0]3]2= 2$[[[0]3 = 2$[[0]3[0]3]

3. No, 2$[0][0] = 2$0][0][0][0 = 2$0][0][0]2] = 2$[[0][0][[0] = 2$0][0][2 = 2$0][0][1][1 = 2$0][0][1][0][0 = 2$[[0][0][1][0]2] = 2$[[0][0][1][0]1][[0][0][1][0]1]

2$[1][0] > 2$0][0][1][0]1][[0][0 = 2$0][0][1][0]1][0][0][0]2 = 2$[[0][0][1][0]1][0][0][[0 = 2$[[0][0][1][0]1][0][0][2] = 2$[[0][0][1][0]1][0][0][1][1] = 2$[[0][0][1][0]1][0][0][1][0][0] = 2$[[0][0][1][0]1][0][0][1][0]2 = 4$[[0][0][1][0]1][0][0][1][0] = 4$[[0][0][1][0]1][0][0][1]4 = 8$[[0][0][1][0]1][0][0][1] = 8$[[0][0][1][0]1][0][0][0][0][0][0][0][0][0][0] = 2048$[[0][0][1][0]1][0][0] = 2048$[[0][0][1][0]1][0]2048

Note that 2$[1][0] would be much larger than 2048$[[0][0][1][0]1][0]2048

About FGH comparisons, if my notation worked as you tought, the notation would be indeed much weaker.