The proof you've provided is essentially showing that the function $f(n) = n + 1$ is one-to-one (or injective). Let's break it down clearly:

Statement to prove:

We want to show that the function $f(n) = n + 1$ is one-to-one. That is, if $f(a) = f(b)$, then we must show $a = b$.

Proof:

1. Assume $f(a) = f(b)$:
   This is the starting assumption. So, we begin by assuming that the function values for two inputs, $a$ and $b$, are equal. That is: $f(a) = f(b)$

2. Substitute the definition of $f$:
   The function is defined as $f(n) = n + 1$, so substitute this into the equation: $a + 1 = b + 1$

3. Simplify:
   Now, subtract 1 from both sides of the equation: $a = b$

4. Conclusion:
   Since we have shown that $f(a) = f(b)$ leads to $a = b$, this confirms that the function $f(n) = n + 1$ is indeed one-to-one (injective).

Thus, $f(n) = n + 1$ is a one-to-one function.

Would you like a further explanation or breakdown of any specific step? Let me know if you have more questions.

Here are 5 follow-up questions you might find useful:

1. What other properties must a function satisfy to be bijective?
2. How do you test whether a function is injective using its graph?
3. Can a quadratic function be one-to-one? Why or why not?
4. What’s the difference between injective and surjective functions?
5. Can a function be injective but not surjective? Can you give an example?

Tip: When trying to prove that a function is one-to-one, it's always useful to start with the assumption $f(a) = f(b)$ and then manipulate the equation to show that it must lead to $a = b$.