proof format updates using gfm

content/CSE442T/CSE442T_L10.md

@@ -86,7 +86,10 @@ f(N,e):\mathbb{Z}_N^*\to \mathbb{Z}_N^*
 $$
 is a bijection.
 
-Proof: Suppose $x_1^e\equiv x_2^e\mod n$
+<details>
+<summary>Proof</summary>
+
+Suppose $x_1^e\equiv x_2^e\mod n$
 
 Then let $d=e^{-1}\mod \phi(N)$ (exists b/c $e\in\phi(N)^*$)
 
@@ -98,13 +101,14 @@ $x_1\equiv x_2\mod N$
 
 So it's one-to-one.
 
-QED
+</details>
 
 Let $y\in \mathbb{Z}_N^*$, letting $x=y^d\mod N$, where $d\equiv e^{-1}\mod \phi(N)$
 
 $x^e\equiv (y^d)^e \equiv y\mod n$
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 It's easy to sample from $I$:
 
@@ -130,7 +134,7 @@ By RSA assumption
 
 The second equality follows because for any finite $D$ and bijection $f:D\to D$, sampling $y\in D$ directly is equivalent to sampling $x\gets D$, then computing $y=f(x)$.
 
-QED
+</details>
 
 #### Theorem If inverting RSA is hard, then factoring is hard.
 

content/CSE442T/CSE442T_L12.md

@@ -82,7 +82,10 @@ The NBT(Next bit test) is complete.
 
 If $\{X_n\}$ on $\{0,1\}^{l(n)}$ passes NBT, then it's pseudorandom.
 
-Ideas of proof: full proof is on the text.
+<details>
+<summary>Ideas of proof</summary>
+
+Full proof is on the text.
 
 Our idea is that we want to create $H^{l(n)}_n=\{X_n\}$ and $H^0_n=\{U_{l(n)}\}$
 
@@ -119,7 +122,7 @@ $\mathcal{D}$ can distinguish $x_{i+1}$ from a truly random $U_{i+1}$, knowing t
 
 So $\mathcal{D}$ can predict $x_{i+1}$ from $x_1\dots x_i$ (contradicting with that $X$ passes NBT)
 
-QED
+</details>
 
 ## Pseudorandom Generator
 

content/CSE442T/CSE442T_L15.md

@@ -115,7 +115,8 @@ $$
 
 #### Theorem PRG exists then PRF family exists.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Let $g:\{0,1\}^n\to \{0,1\}^{2n}$ be a PRG.
 
@@ -184,6 +185,6 @@ Assume that $D$ distinguishes $f_s$ and $F\gets RF_n$ with non-negligible probab
 
 By hybrid argument, there exists a hybrid $H_i$ such that $D$ distinguishes $H_i$ and $H_{i+1}$ with non-negligible probability.
 
-For $H_0$, 
+For $H_0$, $D$ distinguishes $H_0$ and $H_1$ with non-negligible probability.
 
-QED
+</details>

content/CSE442T/CSE442T_L4.md

@@ -88,7 +88,8 @@ $$
 
 is a strong one-way function.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 1. Since $\exist P.P.T.$ that computes $f(x),\forall x$ we use this $q(n)$ polynomial times to compute $g$.
 2. (Idea) $a$ has to succeed in inverting $f$ all $q(n)$ times.
@@ -98,7 +99,7 @@ Proof:
 
     Then $P[a$ inverting $g]\sim P[a$ inverts $f$ all $q(n)]$ times. $<(1-\frac{1}{p(n)})^{q(n)}=(1-\frac{1}{p(n)})^{np(n)}<(e^{-\frac{1}{p(n)}})^{np(n)}=e^{-n}$ which is negligible function.
 
-QED
+</details>
 
 _we can always force the adversary to invert the weak one-way function for polynomial time to reach the property of strong one-way function_