proof format updates using gfm

content/CSE347/CSE347_L9.md

@@ -260,7 +260,8 @@ $$
 
 Claim: the solution to this recurrence is $E[T(n)]=O(n\log n)$ or $T(n)=c'n\log n+1$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 We prove by induction.
 
@@ -296,10 +297,13 @@ If $c'\geq 8c$, then $T(n)\leq c'n\log n+1$.
 
 $E[T(n)]\leq c'n\log n+1=O(n\log n)$
 
-QED
+</details>
 
 A more elegant proof:
 
+<details>
+<summary>Proof</summary>
+
 Let $X_{ij}$ be an indicator random variable that is $1$ if element of rank $i$ is compared to element of rank $j$.
 
 Running time: $$X=\sum_{i=0}^{n-2}\sum_{j=i+1}^{n-1}X_{ij}$$
@@ -344,6 +348,5 @@ E[X]&=\sum_{i=0}^{n-2}\sum_{j=i+1}^{n-1}\frac{2}{j-i+1}\\
 \end{aligned}
 $$
 
-
-QED
+</details>