Sorry? Take a second to think about what you're saying.
In what conceivable sense is "doing a logical and" one thing while "doing a logical nand" isn't?
You can equally claim that
x & y
(x ↑ y) ↑ (x ↑ y)What did you think I meant by this question?
>> In what conceivable sense is "doing a logical and" one thing while "doing a logical nand" isn't?
NAND is more complex than AND, in the sense that it is more expressive than AND (having functional completeness which AND does not).
Similarly, it can be built from other less complex operators (AND and NAND).
If you're taking "One thing" to the extreme, in terms of the granularity or complexity of that "one thing", NAND is not as granular or simple as AND - and therefore isn't taking it to as far "to the extreme".
+---+---+-------+
| a | b | a ↑ b |
+---+---+-------+
| 0 | 0 | 1 |
+---+---+-------+
| 0 | 1 | 1 |
+---+---+-------+
| 1 | 0 | 1 |
+---+---+-------+
| 1 | 1 | 0 |
+---+---+-------+
+---+---+-------+
| a | b | a & b |
+---+---+-------+
| 0 | 0 | 0 |
+---+---+-------+
| 0 | 1 | 0 |
+---+---+-------+
| 1 | 0 | 0 |
+---+---+-------+
| 1 | 1 | 1 |
+---+---+-------+
> It seems naively obvious to me that a(b(x)) is more complex than b(x).
This is just obvious gibberish; if you define b(x, y) = x & y and a(x) = ~x, then you can say "I think a(b(x, y)) looks more complex than b(x, y)", but how do you respond to "when c(x, y) = x ↑ y, I think c(c(x,y), c(x,y)) looks more complex than c(x,y)"? The two claims can't both be true!
Everything, no matter how simple, can be described as the end of an arbitrarily long chain of functions. So what?