Question

Prove that for all sets A and B,
A ∪ B^c = B^c ∪ (A ∩ B).

Answer 1

Answer on Question #76360 – Math – Discrete Mathematics

Question

Prove that for all sets $A$ and $B$,

$$A \cup B^{C} = B^{C} \cup (A \cap B).$$

Solution

Let $x \in A \cup B^{C}$. Then $x \in A$ or $x \in B^{C}$.

If $x \in B^{C}$, then $x \in B^{C} \cup (A \cap B)$.

If $x \in A$ and $x \notin B^{C}$, then $x \in A$ and $x \in B$. So, $x \in A \cap B$. Then $x \in B^{C} \cup (A \cap B)$.

Therefore, in each case $x \in B^{C} \cup (A \cap B)$ and so $A \cup B^{C} \subseteq B^{C} \cup (A \cap B)$.

Let $x \in B^{C} \cup (A \cap B)$. Then $x \in B^{C}$ or $x \in A \cap B$.

If $x \in B^{C}$, then $x \in A \cup B^{C}$.

If $x \in A \cap B$, then $x \in A$ and so $x \in A \cup B^{C}$.

Therefore, in each case $x \in A \cup B^{C}$ and so $B^{C} \cup (A \cap B) \subseteq A \cup B^{C}$.

Since $A \cup B^{C} \subseteq B^{C} \cup (A \cap B)$ and $B^{C} \cup (A \cap B) \subseteq A \cup B^{C}$, $A \cup B^{C} = B^{C} \cup (A \cap B)$.

Hence,

$$A \cup B^{C} = B^{C} \cup (A \cap B).$$

Another way:

Let $X$ be an universal set. Then

$$B^{C} \cup (A \cap B) = (B^{C} \cup A) \cap (B^{C} \cup B) = (B^{C} \cup A) \cap X = B^{C} \cup A = A \cup B^{C}.$$

Hence,

$$A \cup B^{C} = B^{C} \cup (A \cap B).$$

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