Show that $A - B = A \cap B^c$ and hence simplify the following using the laws of set algebra.
a) $A \cap (A - B)$
b) $(A - B) \cup (A \cap B)$
c) $(A \cup B) \cup (C \cap A) \cup (A \cap B)^c$
I wasn't at all sure how to show that $A - B = A \cap B^c$, so I have no attempt for that one.
Note that I've mentioned the set laws used for each step, to ensure that I understand what's going on.
a)
$A \cap (A - B)= (A \cap A) - B$ (By the law of distributivity.)
$= A - B$ (By the idempotence.)
$= A - B$ $\ \ \ Q.E.D.$
b)
$(A - B) \cup (A \cap B) = A - (B \cup (A \cap B))$ (By associativity.)
$= A - (B \cup A) \cap (B \cup B)$ (By distributivity.)
$= A - (B \cup A) \cap B$ (By idempotence.)
$= (A - B) \cup (A \cap B)$ (By associativity.)
$= (A - B) \cup (B \cap A)$ (By commutativity.)
$= A - (B \cup B) \cap A$ (By associativity.)
$= A - B \cap A$ (By idempotence.)
$= A - A \cap B$ (By commutativity.)
$= \emptyset \cap B$
$= \emptyset$ (By domination.)
But the solutions say that this should be $B$.
c)
$(A \cup B) \cup (C \cup A) \cup (A \cap B) = (A \cup B) \cup (C \cap A) \cup (A \cap B)^c = (A \cup B) \cup (C \cap A) \cup A^c \cup B^c$
$= (A \cup B) \cup (C \cup A^c) \cap (A \cup A^c) \cup B^c$ (By De Morgan's law.)
$= (A \cup B) \cup (C \cup A^c) \cap (A \cup A^c)$ (By distributivity.)
$= (A \cup B) \cup (C \cup A^c) \cap U \cup B^c$ (By union with complement.)
$= (A \cup B) \cup (C \cup A^c) \cap U$ (By domination.)
$= (A \cup B) \cup ((C \cup U) \cup (A^c \cap U))$ (By distributivity.)
$= (A \cup B) \cup (C \cup A^c)$ (By identity laws.)
$= (A \cup B \cup C) \cup A^c$ (By associativity.)
$= U$ (Assuming $A$, $B$, $C$ are the only sets in our universe.) $\ \ \ Q.E.D.$
I would greatly appreciate it if people could please take the time to review my work and demonstrate the parts that I wasn't able to do.