Consider the statement $$\text { For all sets } A \text { and } B , ( A - B ) \cap B = \emptyset \text {. }$$
Complete the proof begun below in which the given statement is derived algebraically from
the properties listed in Theorem 6.2.2. Be sure to give a reason for every step that exactly justiﬁes what was done in the step: Proof:
Let $A$ and $B$ be any sets. Then the left-hand side of the equation to be shown is

which is the right-hand side of the equation to be shown. [Hence the given statement is true.]
(The number of lines in the outline shown above works for one version of a proof. If you write
a proof using more or fewer lines, be sure to follow the given format, supplying a reason for
every step that exactly justiﬁes what was done in the step.)

Question

Consider the statement $$\text { For all sets } A \text { and } B , ( A - B ) \cap B = \emptyset \text {. }$$ 
Complete the proof begun below in which the given statement is derived algebraically from
the properties listed in Theorem 6.2.2. Be sure to give a reason for every step that exactly justiﬁes what was done in the step: Proof:
Let $A$ and $B$ be any sets. Then the left-hand side of the equation to be shown is

which is the right-hand side of the equation to be shown. [Hence the given statement is true.]
(The number of lines in the outline shown above works for one version of a proof. If you write
a proof using more or fewer lines, be sure to follow the given format, supplying a reason for
every step that exactly justiﬁes what was done in the step.)

Quizplus · Accepted Answer

The Answer is Proof:
Let

A

and

B

be any sets. Then the left-hand side of the equation to be shown is

\begin{aligned}( A - B ) \cap B & = \left( A \cap B ^ { c } ight) \cap B & ext { by the set difference law } \& = A \cap \left( B ^ { c } \cap B ight) & & ext { by the associative law (for } \cap ) \& = A \cap \left( B \cap B ^ { c } ight) & & ext { by the commutative law (for } \cap ) \& = A \cap \emptyset & & ext { by the complement law (for } \cap ) \& = \emptyset & & ext { by the universal bound law (for } \cap )\end{aligned}

which is the right-hand side of the equation to be shown. [Hence the given statement is true.]

Consider the Statement For all sets A and B,(A−B)∩B=∅. \text { For all sets } A \text { and } B , ( A - B ) \cap B = \emptyset \text {. } For all sets A and B,(A−B)∩B=∅.

Consider the Statement $\text { For all sets } A \text { and } B , ( A - B ) \cap B = \emptyset \text {. }$