Boolean Algebra Simplification - Weird Step When reading the book Boolean Differential Equations(B. Steinbach, C. Posthoff), I stumbled upon the following demonstration of the orthogonality between the simple minimum and the simple derivative:
$$
\underset{x_i}{\min}f(x) \wedge \frac{\partial f(x)}{\partial x_i}=\underset{x_i}{\min}f(x) \wedge (\underset{x_i}{\min} \oplus \underset{x_i}{\max}f(x)))\ \ \ \ (0)
\\
=\underset{x_i}{\min}f(x) \oplus (\underset{x_i}{\min}f(x) \wedge \underset{x_i}{\max}f(x))\ \ \ \ (1)
\\
=\underset{x_i}{\min}f(x) \wedge (1 \oplus \underset{x_i}{\max}f(x))\ \ \ \ (2)
\\
=\underset{x_i}{\min}f(x) \wedge \overline{\underset{x_i}{\max}f(x)}\ \ \ \ (3)
\\
= 0
$$
Step (1) still makes sense to me, you apply the Distributive Law then the Idempotent Law, straightforward. However, the second step eludes me, what law was applied in it?
And wouldn't it be simpler to just do this(starting from (1))
$$
=(\underset{x_i}{\min}f(x) \oplus \underset{x_i}{\min}f(x)) \wedge (\underset{x_i}{\min}f(x) \oplus 
 \underset{x_i}{\max}f(x)) 
\\
=0 \wedge (\underset{x_i}{\min}f(x) \oplus 
 \underset{x_i}{\max}f(x)) 
\\
=0
$$
anyway? Am I missing something very obvious?
 A: Ok, I was missing something very obvious, the applied step is just the [reversed] Distributive Law.
$$
=\underset{x_i}{min}f(x) \oplus (\underset{x_i}{min}f(x) \wedge \underset{x_i}{max}f(x))\ \ \ \ (1)
\\
=\underset{x_i}{min}f(x) \wedge 1 \oplus \underset{x_i}{min}f(x) \wedge (\underset{x_i}{max}f(x))\ \ \ \ (1.1)
\\
=\underset{x_i}{min}f(x) \wedge (1 \oplus \underset{x_i}{max}f(x))\ \ \ \ (2)
\\
(...)
$$
I am forced to do it this way because you cannot apply the Distributive Law, as I tried, for the XOR operation, I was "over-remembering" the Distributive Law example in a previous chapter, namely: 
$$
x_1 \wedge (x_2 \oplus x_3)=(x_1 \wedge x_2) \oplus (x_1 \wedge x_3)
$$
The correct way to simplify (1) would be using De Morgan's Laws and the XOR equivalence, which is essentially what we are doing in (1)-->(3).
$$
A \oplus (A \wedge B) = A \overline{(A \wedge B)} \vee \overline{A}(A \wedge B)
\\
=A(\overline{A} \vee \overline{B}) \vee \overline{A}(A \wedge B)
\\
=(A\overline{A} \vee A\overline{B}) \vee (\overline{A}A \wedge \overline{A}B)
\\
= 0 \vee A\overline{B} \vee (0 \wedge \overline{A}{B})
\\
= A \overline{B}
$$
