Induction

• Induction is a type of proof we can do on recursively defined functions and sets
• Say we are trying to prove \(R(x)\) holds in a recursively defined set \(S = \{S_0, S_1, S_2, \ldots \}\)
• We can prove this by:
  1. Showing \(R(x)\) holds for the base case(s) of \(S\)
  2. Assuming \(R(k)\) holds for all \(k < n\) in the recursive rule, showing that it also holds for step \(n\). In other words, we’re showing \(\big(R(S_0) \land R(S_1) \land \ldots \land R(S_{n-2}) \land R(S_{n-1})\big) \implies R(S_{n})\).
     
• Remember that many things can be defined recursively, so even though \(x \in S\), \(x\) isn’t necessarily a single element. \(x\) can also be a set/function/mapping etc! Think about our recursive powersets definition.

Recall the Fibonacci series…

We want to prove the following about the sum of the first \(n\) numbers of the series:

\(\forall n \in \mathbb{N}, F(0) + F(1) + F(2) + \ldots + F(n-1) + F(n) = F(n+2) - 1\).

S1. State the ‘for all’ statement that you want to prove:

• \(\forall n \in \mathbb{N}, \sum_{i=0}^n F(i) = F(n+2) - 1\).

S2. Say “we prove this by induction on” and state the induction parameter.

• We prove this by induction on \(n\).

S3. Prove the base case(s).

• \(n=0\):
  • \(F(0) = F(2) - 1\)
  • \(0 = 0\) \(\square\)
• \(n = 1\):
  • \(F(0) + F(1) = F(3) - 1\)
  • \(1 = 2 - 1\) \(\square\)

S4. Write Induction Step.

• For a given \(n > 1\),

S5. State the Induction Hypothesis (IH)

• I can assume for all \(1 \leq k \leq n\) that \(\sum_{i=0}^k F(i) = F(k+2) - 1\),

S6. State what you are going to prove about your specific value of \(x\) that was given to you in S4:

• and I want to prove \(\sum_{i=0}^n F(i) = F(n+2) - 1\).

S7. Do the proof for the specific \(x\).

\[\begin{equation*} \begin{array}{l l l} 1. & \forall 1 \leq k \leq n, {} \sum_{i=0}^k F(i) = F(k+2) - 1 & \textrm{IH} \\ 2. & \textrm{Let } k = n - 1, \textrm{ then } \sum_{i=0}^{n-1} F(i) = F(n-1+2) - 1 & \textrm{Applied IH} \\ 3. & \sum_{i=0}^{n-1} F(i) = F(n+1) - 1 & \textrm{Rewrote Line 2}\\ 4. & \sum_{i=0}^{n-1} F(i) + F(n) = F(n+1) - 1 + F(n) & \textrm{Added } F(n) \textrm{ to both sides.}\\ 5. & F(n+2) = F(n+1) + F(n) & \textrm{Fibonacci Def.}\\ 6. & \sum_{i=0}^{n-1} F(i) + F(n) = F(n+2) - 1 & \textrm{Plugged 5. into 4.}\\ 7. & \sum_{i=0}^n F(i) = \sum_{i=0}^{n-1} F(i) + F(n) & \textrm{Def. of Sum} \\ 8. & \sum_{i=0}^{n} F(i) = F(n+2) - 1 & \textrm{Plugged 7. into 6.} \square \end{array} \end{equation*}\]

S8. Declare victory.

• Therefore, we have proved \(\forall n \in \mathbb{N}, F(0) + F(1) + F(2) + \ldots + F(n-1) + F(n) = F(n+2) - 1\).

• Your Inductive Hypothesis (Step 5) should be line 1 in your proof.
• The recursive definition of any structures you’re using (e.g. sum, factorial, exponents, etc.) should be the next lines of your proof.
• Step 6 is what the last line of your proof should be.
• To start, think about how you can use your inductive hypothesis and recursive definitions to get to step 6. (Usually this means plugging a smaller version of \(x\) into your inductive hypothesis and using your recursive definitions to rewrite it.)