Let’s solve the given problem step by step:

a) Montrer que :

(1) $(A \cap B) \cup B^c = A \cup B^c$

Proof:

• By distributive property of union and intersection: $(A \cap B) \cup B^c = (A \cup B^c) \cap (B \cup B^c)$

• $B \cup B^c = E$ (Universal set property), so: $(A \cap B) \cup B^c = A \cup B^c$

(2) $(A - B) - C = A - (B \cup C)$

Proof:

• $A - B = A \cap B^c$, so: $(A - B) - C = (A \cap B^c) \cap C^c$

• By distributive property: $(A \cap B^c) \cap C^c = A \cap (B^c \cap C^c)$

• Using De Morgan's law: $B^c \cap C^c = (B \cup C)^c$

• So: $(A - B) - C = A \cap (B \cup C)^c = A - (B \cup C)$

(3) $A - (B \cap C) = (A - B) \cup (A - C)$

Proof:

• $A - (B \cap C) = A \cap (B \cap C)^c$

• By De Morgan's law: $(B \cap C)^c = B^c \cup C^c$

• So: $A - (B \cap C) = A \cap (B^c \cup C^c)$

• By distributive property: $A \cap (B^c \cup C^c) = (A \cap B^c) \cup (A \cap C^c)$

• $A \cap B^c = A - B$ and $A \cap C^c = A - C$, so: $A - (B \cap C) = (A - B) \cup (A - C)$

b) Simplifier :

(1) $(A \cup B) \cup (C \cup A)$

Simplification:

• By associative property of union: $(A \cup B) \cup (C \cup A) = A \cup B \cup C$

(2) $(A \cap B) \cup (C \cap A)$

Simplification:

• By distributive property: $(A \cap B) \cup (C \cap A) = A \cap (B \cup C)$

Summary:

1. $(A \cap B) \cup B^c = A \cup B^c$
2. $(A - B) - C = A - (B \cup C)$
3. $A - (B \cap C) = (A - B) \cup (A - C)$
4. $(A \cup B) \cup (C \cup A) = A \cup B \cup C$
5. $(A \cap B) \cup (C \cap A) = A \cap (B \cup C)$

Would you like detailed verification for any specific part?

Relative Questions:

1. How can De Morgan's laws simplify set operations?
2. What is the importance of distributive properties in set theory proofs?
3. How can universal sets $E$ simplify logical operations?
4. What are common pitfalls when applying set subtraction rules?
5. How are these properties applied in real-world problem solving?

Tip: Always double-check your steps using Venn diagrams to visualize set relationships clearly.