Additive Function is Strongly Additive/Proof 2

Proof
Recall that $\sqcup$ denotes the disjoint union.

First, since:
 * $A \cup B = A \sqcup \paren {B \setminus A}$

we have:
 * $\paren 1 : \quad \map f {A \cup B} = \map f A + \map f {B \setminus A}$

Secondly, since by Set Difference and Intersection are Disjoint and Set Difference Union Intersection:
 * $B = \paren {B \setminus A} \sqcup \paren {A \cap B}$

we have:
 * $\paren 2 : \quad \map f B = \map f {B \setminus A} + \map f {A \cap B}$

On the other hand, from $A \cap B \subseteq A \cup B$ follows:
 * $A \cup B = D \sqcup \paren {A \cap B}$

where:
 * $D := \paren {A \cup B} \setminus \paren {A \cap B}$

so that:
 * $\map f {A \cup B} = \map f D + \map f {A \cap B}$

In particular, we have neither:
 * $\map f {A \cup B} = + \infty \land \map f {A \cap B} = - \infty$

nor:
 * $\map f {A \cup B} = - \infty \land \map f {A \cap B} = + \infty$

Thus:
 * $\map f {A \cup B} + \map f {A \cap B}$

is well-defined.

Finally: