Why are $\lambda$-systems defined with the stronger condition $A\subset B\implies B\setminus A\in\Lambda$

Question

A class $\Lambda$ over subsets of $E$ is called a $\lambda$-system if among other conditions the following is fullfilled:

$$(A,B\in\Lambda) \wedge (A\subset B) \implies B\setminus A\in\Lambda$$

My question is if there is a reason why the antecedent of this conditional requires that $A\subset B$. If we would omit this and define a $\lambda$-system as a A class $\Lambda$ s.t. (among the other two conditions):

$$(A,B\in\Lambda) \implies B\setminus A\in\Lambda$$

Since $(A,B\in\Lambda) \wedge (A\subset B) \implies (A,B\in\Lambda)$, the latter definition would have a weaker antecedent in its conditional. We would also still have that every $\sigma$-algebra is a $\lambda$-system, so what, if there exists any, is the reason to define it like this?

Answer 1

If one weakens the condition as said, that is considers sets $\Lambda \subseteq \def\P{\mathfrak P}\P(S)$ that have

(1) $S \in \Lambda$
(2) If $A, B \in \Lambda$, then $A \setminus B \in \Lambda$
(3) If $(A_n) \in {}^{\mathbf N}\Lambda$ is increasing, then $\bigcup_n A_n \in \Lambda$.

then every $\Lambda$ satisfying (1)-(3) is - of course - a $\lambda$-system, but one has strenghed the conditions, so there may be $\lambda$-systems not satisfying (1)-(3). And there are: Any $\lambda$-system satisfying (2) is a $\pi$-system, due to $$ A \cap B = A \bigm\backslash (A \setminus B) $$ Hence, any $\Lambda$ satisfying (1)-(3) is in fact a $\sigma$-algebra. Hence (1)-(3) do characterize $\sigma$-algebras. So, if we want to study $\lambda$-systems in their own right, we must stuck with

(1) $S \in \Lambda$
(2') If $A, B \in \Lambda$ with $B \subseteq A$, then $A \setminus B \in \Lambda$
(3) If $(A_n) \in {}^{\mathbf N}\Lambda$ is increasing, then $\bigcup_n A_n \in \Lambda$. or, equivalenty (1) $S \in \Lambda$
(2'') If $B \in \Lambda$, then $S \setminus B \in \Lambda$
(3') If $(A_n) \in {}^{\mathbf N}\Lambda$ is a pairwise disjoint sequence, then $\bigcup_n A_n \in \Lambda$.