Prove $\bf{H}(\bf{x} + \bf{y}) = 0$ iff $\bf{Hx}=\bf{Hy}$,

Question

where $\bf{H}$ denotes the parity-check matrix of a code in $\mathbb{B}^n$ and $\bf{x}$ and $\bf{y}$ are any two words in $\mathbb{B}^n$.

Comment

Note: if $\bf{H}(\bf{x}+\bf{y})=0$, then $(\bf{x}+\bf{y})$ is a codeword.

Not sure how to proceed beyond the proof I have below.

Proof

Suppose $\bf{H}x = \bf{H}y$, then $\bf{H}(x+y) = \bf{H}x + \bf{H}y = \bf{H}x + \bf{H}x = 2\bf{H}x$

Answer 1

In any field, $H(x+y)=0$ if and only if $Hx+Hy=0$ if and only if $Hx=-Hy$. But in $\Bbb B^n$, $-Hy=Hy$, because the addition is $\Bbb B$ is defined by $1+1=0$.

Answer 2

So your first part of the proof is correct, you just need to add $2Hx=0$ because you are working modulo 2. Notice $$H(x+y)=Hx+Hy=0$$ so $Hx=-Hy=Hy$, this is again because you are working in $\mathbb{B}=\mathbb{F}_2$, so $-1=1$.

Also the step $H(x+y)=Hx+Hy$ is allowed because matrices with coefficients in $\mathbb{B}$ are a ring.