Why does this regularity still match the second space?

[dublove]

I was trying to avoid starting with three consecutive spaces.
I stumbled upon this expression not working.
That is, I was trying to avoid the “what” line.

Is my regularity wrong?

[Eugene Tyson]

It works fine for me

Don't know why it's so complicated - you could search for a single space (inserted by space bar) and it finds the same thing.

Can you elaborate or provide a larger sample set?

[Robert at ID-Tasker]

@dublove

What exactly are you trying to achieve? 

Remove those spaces? All of them or some? 

Replace them with something else? 

Remove whole line? 

[Peter Kahrel]

To avvoid lines that start with three spaces, look for

^[^\h][^\h][^\h]

[FRIdNGE]

^\S{3}

(^/)  The Jedi

[dublove]

@Peter Kahrel 

I would like to use a forward and reverse check(? = or ? <!) to realize it。

Well, a different example might make more sense.
I'm trying to find and replace a space.
Except for spaces after numbers, except for consecutive spaces at the beginning of a paragraph.

I just realized that I can't replace spaces in (? <! [\d])(\h) expression, then add the condition.
(? <! [\d\h])(\h).

That is to say: (? <! [\h])(\h) may be meaningless.

[Robert at ID-Tasker]

@dublove

If you add "^" at the beginning of your GREP expression - it will find only paragraphs starting with your search phrase - as per examples from @Peter Kahrel and @FRIdNGE. 

[FRIdNGE]

You write:

"I'm trying to find and replace a space."

What space in your sample?

Apparently, you want to find a space not preceded by a number or another space, the "3 spaces" case being not taken in account. Right?

(^/)

Between, "(?<![\d\h])" won't never work because it's a Grep incorrect syntax.

[dublove]

Theoretically there is no problem.
In practice:(? <! \h)([a-z]) is established
But (? <! \h)(\h) does not hold.