 We will next prove that:
 There is a matching
of n men (to
n women)
if:
 For every subset of
m men
(m ≤ n),
the m men
know at least m women


Rewritten:
 If
every subset of
m men
(m ≤ n),
these m men
know at least m women,
then:
 There is a matching
of n men (to
n women)


We show this by construction
(it's actually an induction proof)
 Base case of the induction proof:
 We can find a matching for
1 (m = 1) man

Proof:
 Since every subset of
1 men
the 1 men
knows at least 1 women,
we just match
that man with any one of the woman that he knows....

 Suppose that we have found a
matching for
m men and
m < n:
We will prove that:
 There is a matching for
m+1 men

 Since m < n,
there is at least one man (p) that
is unmatched:
There are 2 possibilities:
 p knows
some q ∉ B
In this case, it is trivial to
prove that there is a
matching for
m+1 men:
 p
only knows
women q ∈ B
The reason that
p
cannot find a match
is:

We will now show how to construct a match for this difficulyt case.
 We construct a
m+1 matching
using:
 p and
 All the elements in the
set A

as follows:
 Start with man p
 Given: for every subset of
m = 1 man
the 1 man (p)
know at least 1 woman
Do this:
 Match
p up
with someone he knows.

 Note:
we know that she is
some element ∈ B


Now, 2 men
(p and q) that
are matched up with the
same woman:
Because we have:
 For every subset of
m = 2 men
the 2 men
know at least 2 woman

There are 2 possibilities:
Case 1: the second woman ∉ B

Case 2 the second woman ∈ B



 In the first case
(when the
other woman ∉ B)
we have found a m+1 matching:
 In the second case
we repeat the argument
using
3 elements:
Because we are given that:
 For every subset of
m = 3 men
the 3 men
know at least 3 woman

There are again 2 possibilities:
Case 1: the second woman ∉ B

Case 2: the second woman ∈ B



 And so on...
 Eventually, we will reach
m+1
Then we will have found a m+1
matching
I.e.,: m+1 men
can be matched with
a different woman that
they know

 Therefore, we have shown that:
 If there is a matching of
m men and
m < n,
we can always find a larger matching of
m+1 men

Therefore:
 There is a matching of
n men

