First, the integer i can be expressed in binary as a1b where

a =  sequence of 0's and 1's.

1 =  the LSB (least-significant bit)

b =  sequence of 0's (possibly empty)

Then i - 1 can be expressed as a0b^c where b^c is the complement of b (a sequence of 1's, possibly empty)

Following this binary representation i & (i - 1) = a1b & a0b^c = a

We can apply the same principle to represent i - (i& - i):

 - i = a^c1b, why? Because  - i is the two-complement of i, that is flipping all the bits (one-complement of i) and adding 1

If we apply one-complement to i we get a^c0b^c, and adding 1 we get a^c1b

Now, i &  - i = a1b & a^c1b = z1b where z is a sequence of zeros with the same length as a

i - (i &  - i) = a1b - z1b = a

So, i - (i &  - i) = i & (i - 1) = a, (that is, turn off the LSB).

Hope this helps :)