#### Step 1: Show that P \Rightarrow Q is equivalent to (P \lor Q) \Leftrightarrow Q
1. Start with P \Rightarrow Q which is logically equivalent to \neg P \lor Q.
2. Take (P \lor Q) \Leftrightarrow Q.
- **Case 1:** If Q is true, both (P \lor Q) and Q are true, so (P \lor Q) \Leftrightarrow Q is true.
- **Case 2:** If Q is false, then P \lor Q must be false. This requires P to be false. If Q is false, then P is also false, consistent with P \Rightarrow Q.
Therefore, P \Rightarrow Q and (P \lor Q) \Leftrightarrow Q are equivalent.
#### Step 2: Show that P \Rightarrow Q is equivalent to (P \land Q) \Leftrightarrow P
1. Start with P \Rightarrow Q which is equivalent to \neg P \lor Q.
2. Consider (P \land Q) \Leftrightarrow P:
- If P is true, P \land Q is true if and only if Q is true, corresponding to P \Rightarrow Q.
- If P is false, both sides of (P \land Q) \Leftrightarrow P are false, which is consistent with P \Rightarrow Q.
Thus, P \Rightarrow Q and (P \land Q) \Leftrightarrow P are equivalent.
Since (P \lor Q) \Leftrightarrow Q and (P \land Q) \Leftrightarrow P are both equivalent to P \Rightarrow Q, all three statements are equivalent.
### Part (b): Show that A \subseteq B if and only if A \cap B = A if and only if A \cup B = B
[Solution]
All three statements are equivalent.
[Step-by-Step]
#### Step 1: Show that A \subseteq B if and only if A \cap B = A
1. A \subseteq B implies for all x \in A, x \in B. Thus, x \in A \cap B, so A \subseteq A \cap B.
2. A \cap B \subseteq A by definition, thus A = A \cap B.
Conversely, A = A \cap B implies any x \in A is also in B, so A \subseteq B.
#### Step 2: Show that A \subseteq B if and only if A \cup B = B
1. A \subseteq B implies all elements of A are in B, thus A \cup B = B.
2. A \cup B = B implies all x \in A are in B, thus A \subseteq B.
### Conclusion:
We have shown that:
- A \subseteq B
- A \cap B = A
- A \cup B = B
These statements are equivalent, proving part (b) of the question.