---
# System prepended metadata

title: Lab 10

---

# Lab 10
## Nick Cardenas and Jeronimo Camargo
### Problem 1: The Other Side
**Claim:** $\overline{A} \cap \overline{B} \subseteq \overline{A \cup B}$
 
*Proof:* Let x $\in$ $\overline{A} \cap  \overline{B}$ 
Consider whether $x \in \overline{A}$, by definiton of compliment we know that $x \not \in  A$. By definition of intersection, we know that $x$ is $x \not \in  B$, meaning  that the intersection of set $\overline{A}$ and $\overline{B}$ includes $x$. So, we can conclude that the union of this two sets will indeed include $x$. 

### Problem 2: Pivot

1. $S=$ {$1,2,3,4$}
    Pivot $a = 3$
    $T_1$ = $P(S-a)$ = 
    
    { $\emptyset$
      {1},{2},{4}
      {1,2} {1,4},{2,4}
      {1,2,4}
    }
    
    $T_2$ = {1,2,3,4} To get subset $B$, we take one of the sub-sets generated in  $T_1$
    
    Our concrete $T_1$ and $T_2$ work as a partition because the interception of both sets results in an empty set and its union is part of the $P(S)$
    
2. **Claim:** $P(S) \subseteq T_1 \cup T_2$
$T_1$ is the power set of the set resulting in the set difference of set $S$ and the set containing only the element $a$
$T_2$ is a group of sets that contains each set in $T_1$ with and adds the elemenet $a$.
Therefore, when these two sets are combined, as we see in the claim, we get a set of all possible subsets of S, or $P(S)$
    
