Proof that modulo 2 addition is reversed with the same function

Question

This is from a cryptography book description encryption & decryption using stream ciphers

Encryption $ e(x_i) = y_i \equiv x_i + s_i mod 2$

where $x_i, y_i, s_i \in\ \{0,1\} $

Prove that
$x_i \equiv y_i + s_i mod 2$

Proof:

$ e(y_i) \equiv y_i + s_i mod2$
subsitute value of $y_i$ in the above

$e(y_i) \equiv x_i + s_i + s_i mod 2$ ---- (1)

$e(y_i) \equiv x_i + 2 s_i mod 2$

$e(y_i) \equiv x_i + 0 mod 2$

$e(y_i) \equiv x_i mod 2$ Q.E.D* --- (2)

---

I didn't understand 2 things here (1) & (2)

(1) Shouldn't (1) have been $e(y_i) \equiv x_i + s_i mod 2 + s_i mod 2$ instead of $e(y_i) \equiv x_i + s_i + s_i mod 2$

(2) How do they arrive at the Q.E.D

All that was prove in (2) was that $e(y_i) \equiv x_i mod 2$

What was required to be proven was $e(y_i) \equiv x_i + s_i mod 2$

So how is the proof completed?

Answer 1

You are given that $y_i=e(x_i)=(x_i+s_i) \pmod{2}$. Here $x_i, s_i \in \{0,1\}$ (bits) and the operation $\mod 2$ is applied to the sum of the bits $x_i$ and $s_i$.

Now we want the decryption function $d$ so that $d(y_i)=x_i$. The claim is that the encryption and decryption functions are the same, i.e. $x_i=d(y_i)=(y_i+s_i) \pmod{2}$.

So, \begin{align*} d(y_i)&=(y_i+s_i) \pmod{2}\\ &=(\overbrace{x_i+s_i}^{y_i}+s_i) \pmod{2} & \because s_i = s_i \bmod{2}\\ &=(x_i+2s_i) \pmod{2}\\ &=x_i \pmod{2}\\ &=x_i & \because x_i = x_i \bmod{2} \end{align*}

Answer 2

For your first doubt: Let $\bar{a} = a \pmod{p}$ and $\bar{b} = b \pmod{p}$ where $p \in \mathbb{P}:=$ set of all primes. Then we define $$\overline{a + b} := (a+ b)\pmod{p} \equiv a \pmod{p} + b \pmod{p} = \bar{a} + \bar{b} $$ Thus, $ x_i + s_i + s_i \pmod{2} \equiv x_i + s_i \pmod{2} + s_i \pmod{2}$, where $x_i, s_i \in \mathbb{Z}/2\mathbb{Z} := \{ 0, 1\}$