Determine the form of the following statement:

Integer n is even if and only if 5n^2+6n+7 is odd.

Determine the form of the following statement:

The product of two odd integers is odd.

Determine the form of the following statement:

3n -6 = 9 has a single solution.

In the class, the professor explained Chain-of-Thought (CoT), which is a state-of-the-art reasoning model, and he also emphasized the importance of writing a clear proof with detailed statements. 

As a step of learning, you will need to write a table on paper for a detailed proof and CoT. The table shall have three columns:

1. Step (copy the numbers in the proof on Page 3 of "Chapter 1 - Forms and Techniques for Proofs")
2. Statement (copy the statement in the proof on Page 3 of "Chapter 1 - Forms and Techniques for Proofs")
3. CoT: choose 5 most important steps as your CoT for this proof, and specify them by using 1, 2, 3, 4, 5 in this column.

For this problem, you need to scan (or take a photo of) your table, generate a jpg file, and then upload it.

Given the following valid argument, choose the best answer.

1. P(a) \wedge (\forall x~ (P(x) \to Q(x)))

2. \forall x~ (P(x) \to Q(x))

3. P(a) \to Q(a)

4. P(a)

5. Q(a)

Which inference rule is used to derive step 2?

Given the following valid argument, choose the best answer.

1. P(a) \wedge (\forall x~ (P(x) \to Q(x)))

2. \forall x~ (P(x) \to Q(x))

3. P(a) \to Q(a)

4. P(a)

5. Q(a)

Which inference rule is used to derive step 3?

Given the following valid argument, choose the best answer.

1. P(a) \wedge (\forall x~ (P(x) \to Q(x)))

2. \forall x~ (P(x) \to Q(x))

3. P(a) \to Q(a)

4. P(a)

5. Q(a)

Which inference rule is used to derive step 5?

Given the following valid argument, choose the best answer.

1. p \wedge (p \to q)

2. p

3. p \to q

4. q

Which inference rule is used to derive step 2?

Given the following valid argument, choose the best answer.

1. p \wedge (p \to q)

2. p

3. p \to q

4. q

Which inference rule is used to derive step 3?

What is the inference rule for the following logic?

(((\neg p) \wedge (p \vee q)) \to q) \equiv T