Set Difference Union Second Set is Union

Theorem

The union of a set difference with the second set is the union of the two sets.

That is, let $S, T$ be sets.

Then:

$\paren {S \setminus T} \cup T = S \cup T$

Consider $S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universal set.

$\ds \paren {S \setminus T} \cup T$

$=$

$\ds \paren {S \cap \map \complement T} \cup T$

$\ds $

$=$

$\ds \paren {S \cup T} \cap \paren {\map \complement T \cup T}$

$\ds $

$=$

$\ds \paren {S \cup T} \cap \mathbb U$

$\ds $

$=$

$\ds S \cup T$

$\blacksquare$

This is one of the logical axioms that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.

However, the intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom.

This in turn invalidates this theorem from an intuitionistic perspective.

1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $\S 1.1$: Theorem $1.3$
1968: A.N. Kolmogorov and S.V. Fomin: Introductory Real Analysis ... (previous) ... (next): $\S 1$: Sets and Functions: Problem $2$

This problem is worded: Show that in general $A - B \cup B \ne A$.