Show an equivalence via induction Let $f$ be a set function $f: 2^{V} \rightarrow \mathbb{N}_{0}$; let $S,T\subset V$ be such that $S \subset T$ and let $j$ be any element such that $j \in (V \setminus T)$ (so $j$ doesn't belong to $T$). Suppose it holds that:
$f(S \cup j ) - f(S) \ge f(T \cup j ) - f(T) $
($\mathbb{N}_{0}$ is the set of natural numbers including $0$).
it can be proven this above property to be equivalent to the following statement (under the same definition as for $f$):
Let $A,B \subset V$ then the above property implies $f(A \cap B) + f(A \cup B) \le f(A)+f(B) $
I know the beginning of the proof. Exploiting that $(A \cap B) \subset B$, we can inductively apply the inequality to each element of $A \setminus B$ to get the equivalence. 
I do not understand what here is meant by last sentence to get the proof.
Can anyone help me in this?
 A: It's important to note that $V$ is a finite set (or the induction would not work). 
So suppose $f$ satisfies the submodular condition (as the first condition is named in the linked proof).
Note that $A = (A \cap B) \cup (A \setminus B)$. If $A \subseteq B$ (i.e. $A \setminus B = \emptyset$, then $A \cap B = A, A \cup B = B$ and the inequality is an equality. Then suppose $A \setminus B = \{j_0, \ldots,j_n\}$. Then we get:
$$f((A \cap B) \cup \{j_0\}) - f(A \cap B) \ge f(B) - f(B \cup \{j_0\})$$
$$f((A \cap B) \cup \{j_0,j_1\}) - f((A \cap B) \cup \{j_0\}) \ge f(B \cup \{j_0,j_1\}) - f(B \cup \{j_0\})$$
$$\ldots$$
$$  f((A \cap B) \cup \{j_0,j_1,\ldots,j_n\}) - f((A \cap B) \cup \{j_0,j_1,\ldots,j_{n-1}\}) \ge f(B \cup \{j_0,j_1,\ldots,j_n\}) - f(B \cup \{j_0, j_1, \ldots, j_{n-1}\})$$
and adding them all up (everything telescopes except two terms) and using that $$(A \cap B) \cup \{j_0,j_1,\ldots,j_n\} = (A \cap B) \cup (A \setminus B)  = A$$ and 
$$B \cup \{j_0,j_1,\ldots,j_n\} = B \cup (A \setminus B) = A \cup B$$ gives: 
$$f(A) - f(A \cap B) \ge f(A \cup B) - f(B)$$ which is what we need as we can re-arrange terms (just finite numbers).
The induction remark is a way to make this argument more formal (getting rid of the dots). So we prove the statement by induction on the size $n$ of $A \setminus B$ and we already checked the case $n=0$ (the empty case). Suppose that it holds for all pairs $A,B$ with $|A \setminus B| = n$ ($n \ge 0$) and suppose now that $A,B$ are such that $|A \setminus B| = n+1$. Pick any fixed $j \in A \setminus B$. Define $A'= A\setminus \{j\}$. Note that for $A',B$ we have that $|A' \setminus B| = n$ so we can apply the induction hypothesis:
$$f(A' \cap B) + f(A' \cup B) \le f(A') + f(B)\text{.}$$
Note that $A' \cap B = A \cap B$. Also note that $A' \subseteq A' \cup B$ and $j \notin A' \cup B$. So we apply the modular property to get:
$$f(A' \cup \{j\}) - f(A') \ge f(A' \cup B \cup \{j\}) - f(A' \cup B)\text{.}$$
and as $A' \cup \{j\} = A$ and $A' \cup B \cup \{j\} = A \cup B$:
$$f(A) - f(A') \ge f(A \cup B) - f(A' \cup B) \leftrightarrow f(A \cup B) - f(A' \cup B) \le f(A') + f(B)$$ and adding the inequalities from the induction hypothesis and this final one gives us the desired inequality for $A,B$, finishing the induction. It's essentially the same argument, though. It is a bit cleaner.  
For the other direction, suppose that $f$ satisfies the condition $$f(A \cap B) + f(A \cup B) \le f(A) + f(B)$$ for all $A,B \subseteq V$. 
Then suppose $S \subseteq T$ and $j \notin V \setminus T$. Then use the property for $A = T$ and $B = S \cup \{j\}$; note  that $A \cap B = T \cap (S \cup \{j\}) = (T \cap S) \cup (T \cap \{j\})= S$, $A \cup B = S \cup T \cup \{j\} = T \cup \{ j \}$, so we have 
$$f(S) + f(T \cup \{j\}) \le f(T) + f(S \cup \{j\})$$ and we re-arrange this to 
$$f(T \cup \{j\}) - f(T) \le f (S \cup \{j\}) - f(S)$$ which is exactly the submodular condition.
