I need help understanding how this solution was made:
$(A'∩B')∪A=A∪B'$
$[(A')'∪B]∪A=A∪B'$
$(A∪B')∪A=A∪B'$
$A∪(A∪B')=A∪B'$
$A∪B'=A∪B'$
I don't really know how our instructor arrived to that answer. How to prove that $(A'∩B')∪A=A∪B'?$
I need help understanding how this solution was made:
$(A'∩B')∪A=A∪B'$
$[(A')'∪B]∪A=A∪B'$
$(A∪B')∪A=A∪B'$
$A∪(A∪B')=A∪B'$
$A∪B'=A∪B'$
I don't really know how our instructor arrived to that answer. How to prove that $(A'∩B')∪A=A∪B'?$
One way to see this is using a Venn diagram
$A$ is the blue and green
$B$ is the blue and white
$A'$ is the yellow and white
$B'$ is the yellow and green
So $A' \cap B'$ is the yellow, and $(A' \cap B') \cup A$ is the yellow, blue and green
while $A \cup B'$ is also the blue, green and yellow, so they are equal.
Another approach is $A \cup B' = A \cup [(A \cap B') \cup (A' \cap B')] = [A \cup (A \cap B')] \cup (A' \cap B') $ $= A \cup (A' \cap B') = (A' \cap B') \cup A$
supposing that A' and B' are the complements of $A$ and $B$ with respect to some set $X$, we have for $x \in X$:
If x \in (A' \cap B') \cup A then if $x \in A$ obviously x \in A \cup B'. If $x \not\in A$, we must have x \in A' \cap B', so x \in B' and therefore x \in A \cup B'.
If x \in A \cup B': If $x \in A$ then x \in (A' \cap B') \cup A, otherwise i. e. if $x \not\in A$ we must have x \in B'. As $x \not\in A$ we have x \in A' and therefore x \in A' \cap B' \subseteq (A' \cap B') \cup A.
Now we have shown (A' \cap B') \cup A \subseteq B' \cup A and B' \cup A \subseteq (A' \cap B') \cup A. So the two sets are equal.