The problem goes as follows:
Define the intersection of binary vectors $x,y$ to be the vector: $x*y = (x_1y_1,\ldots , x_n y_n ) $ which has 1's only where both $x$ and $y$ do. Show that $$wt(x+y) = wt(x) + wt(y)-2wt(x* y).$$ Where $wt$ is Hamming weight, defined as the number of non zero entries in the vector.
I want to show this identity then I thought of the following proof, let me know if it's ok or need some tweaking.
$$wt(x+y) = \# \text{of non 0's in (x+y)}=\#\text{of non zero in x and zero in y}+\#\text{of zero in x and non-zero in y}=wt(x)-wt(x*y)+wt(y)-wt(x*y)$$
Now obviously they are symmetric so I thought to myself it's enough to prove for the first term above.
If we subtract from $\# \text{of non zeros in x}$, $wt(x*y)$, we get $\# \text{of non zero in x and zero in y}$. If there's a $0$ in $y$ and a nonzero in $x$ then subtract from $wt(x)$ the $\# \text{ of nonzeros in x the # of nonzeros in $x*y$}$. And this justifies the claim.
Is this right, or there's another subtler justification?
Thanks.
Perhaps I'm just not following your argument, but it doesn't look to me that you've actually justified why $$\#\text{entries that are 1 in x and 0 in y}=wt(x)-wt(x*y).$$ Perhaps it's easier to show $$wt(x)=\#\text{entries that are 1 in x and 0 in y}+wt(x*y).$$ The left hand side is the size of the set $A=\{i:x_i=1, y_i =0\text{ or }1\}$. The first term on the right hand side is the size of the set $B=\{i:x_i=1,y_i=0\}$, and the last term on the right hand side is the size of the set $C=\{i:x_iy_i=1\}=\{i:x_i=1,y_i=1\}$. Now it should be clear that $B$ and $C$ are disjoint and $A=B\cup C$, so it follows that $|A|=|B|+|C|$, and this proves the claim.