Proof verification: Commutativity of set union and intersection Theorem. Let $M, N, L$ be sets. Then the following assertions hold:
(1) $M \cap N = N \cap M$
(2) $M \cup N = N \cup M$
Proof is left as an exercise.

My attempt:
(1) $\forall x: x \in M \cap N \implies x \in M \land x \in N \implies x \in N \cap M \implies N \cap M $
(2) $\forall x : x \in M \cup N \implies x \in M \lor x \in N \implies x \in N \cup M \implies N \cup M$
Is this right or am I either being redundant or jumping to conclusions?

Liesen, J., Mehrmann, V. 2015. Linear Algebra. Berlin, Germany.: Springer.
 A: Showing set equality (often) comes down to showing that each is a subset of the other. I’ll show one direction. You can prove the other directions.
1) Let $x\in M\cap N$. Then $x\in M,N$. Then $x\in N\cap M$. Thus, $M\cap N \subset N\cap M$.
2) Let $x\in M\cup N$. Then $x\in M$ or $x\in N$. Then $x\in N\cup M$. Thus, $M\cup N \subset N\cup M$.
A: I am proving this using natural deduction:


*

*x                   (arbitrary constant - a.k.a - let there be)

*$x\in M \cup N$         (let's assume)

*$x\in M \vee x \in N$    (Def of union, 2)

*$ x\in N \vee x\in M$       (communitativity of disjunction, 3)

*$ x\ N \cup M $             (Def of union, 4)

*$x\in M \cup N \rightarrow x\in N \cup M$     (Introduction of implication, 2 to 5)

*$x\in N \cup M$         (let's assume)

*$x\in N \vee x \in M$    (Def of union, 2)

*$ x\in M \vee x\in N$       (communitativity of disjunction, 3)

*$ x\ M \cup N $             (Def of union, 4)

*$x\in N \cup M \rightarrow x\in M \cup N$     (Introduction of implication, 7 to 10)

*$x\in M \cup N \leftrightarrow x\in N \cup M $  (Introduction of equivalence, 11,6)

*$\forall M,N.(x\in M \cup N \leftrightarrow x\in N \cup M)$  (Introduction of universal, 1-12)
QED


It could be written a lot nicer when there's an option to use tabs, and indents.
