Why does subtracting 1 from $-x$ in $y= \sqrt{-x}$, getting $y = \sqrt {-x -1 }$, result in a LEFT horizontal shift? Correct me if I am wrong, but I think that, as a general rule: 


*

*When the input of a function is diminished, that is when $f(x)$ becomes $f( x-n )$ with $(n>0)$ , then the graph is moved to the right.

*When the input of a function is increased, that is when $f(x)$ becomes $f( x+n )$ with $( n>0)$, then the graph is moved to the left. 
It took me some time to understand this, but now it seems to me that I come across an exception, with $f(x) =  \sqrt{ - x -1}$ which moves 1 unit to the left relatively to $f(x)= \sqrt{ - x}$. 
Could someone explain to me what I've missed? 

NOTE : The explanations given below make me realize what was wrong in my question. 
I thought that the case I indicate was an exception to the general rule : 
" f(x-n) ( with n positive) produces a horizontal shift to the right". 
BUT , actually, f(x) = sqrt( -x - 1 ) is NOT a case that falls under this rule. 
In order to have a case falling under the rule, I should have treated ( x-1) as the new input, in order to get : f(x-1) = sqrt [ - ( x-1) ] = sqrt (-x +1). 
The new function f ( now correctly defined) perfectly obeys the rule: with a graphing calculator, I can see that the graph moves 1 unit to the right. 
 A: You have $f(x)=\sqrt{-x}$, and by "subtract $ 1 $ from $ -x $" what you're doing is calculate $f(x+1)$, which moves the original graph of $ f (x) $ one unit to the left. Look
$$f(x+1)=\sqrt{-(x+1)}=\sqrt{-x-1}$$
What you want to do is compute $ f (x-1) $, that is, $ \sqrt{-x + 1} $ so that the graph moves one unit to the right.
A: When the INPUT of  a function is increased/decreased....
Notice you are subracting the $1$ somewhere in the output.
If $f(x) = \sqrt{-x}$ then $f(x+1) = \sqrt{-(x + 1)} = \sqrt{-x -1}$ and $f(x-1) = \sqrt{-(x-1)} = \sqrt{-x+1}$.
So $-1$ in the output does not mean a $-1$ in the input.  In fact it absolution means a $+1$ in the input.
....
Here's another "paradox"  Let $f(x) = 4x$ and $f(x)$ passes through $(0,0)$ but $4x + 1$ which is $+1$ should be a of $1$ to the left, passes through $(0,-\frac 14)$ so the graph is only shifted $\frac 14$ to the left.  How did a $1$ become $\frac 14$.
Again a $+1$ in the ouput doesn't mean a $+1$ in the input.  Noting $f(x+\frac 14) = 4(x +\frac 14) =4x + 1$ , a s $+1$ in the output means a $+\frac 14$ in the input.
