TITLE: Derive Hall's theorem from Tutte's theorem
QUESTION [7 upvotes]: I'm trying to derive:
Hall Theorem A bipartite graph G with partition (A,B) has a matching of A $\Leftrightarrow \forall S\subseteq A, |N(S)|\geq |S|$
From this:
Tutte Theorem A graph G has a 1-factor $\Leftrightarrow \forall S\subseteq V(G), q(G-S)\leq |S|$
where $q()$ denotes the number of odd connected components.
The idea of the proof is to suppose true the Tutte's condition for a bipartite graph G and by contradiction suppose that $\exists S\subseteq A,|N(S)|<|S|$
Now consider what happens if we remove from G the vertices of $N(S)$. All the vertex of $S$ become isolated vertex (odd component), and potentially we also have some other odd component, so we know that:
$|S|\leq q(G-N(S))\leq |N(S)|$
where we apply Tutte's theorem in the second inequality. Is this a valid proof?

REPLY [6 votes]: It’s somewhat easier to prove the converse of the above statement:  the existence of a set $S$ not satisfying the Hall criterion $|S|≤|N(S)|$ implies the existence of a set $T$ not satisfying the Tutte criterion, so that $G−T$ has more than $T$ odd components. So, if $G$ does  not  meet  the  conditions  of  Hall’s  Theorem,  then  there  is  a  vertex-set $S⊂A$ such that $|S|>|N(S)|$.  Let $T=N(S)$.  By the definition of a neighborhood,every element of $S$ is adjacent to an element of $T$, so every element of $S$ is an isolated vertex  of $G−T$.   Thus, $G−T$ contains  at  least $|S|$ isolated  vertices,  which  are  in themselves odd components; since $|S|>|T|$, this set violates Tutte’s Theorem.
#Note that this is just a proof from this link with sole purpose to benefit the readers because I believe this solution is somewhat more 'intuitive' and simpler.