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.