Prove $p_1 \iff p_2 \iff p_3 \iff \ldots \iff p_n$ by using Proof by Induction.

Question

I had to write a Python program that shows the proper output of an "If and only If" truth table. My program works, and now I have to use induction that my formula is correct for all positive integers. We studied proof by induction a few weeks ago, but by no means is it easy to me. I'm assuming that I have to do something like proving $p$ implies $q$, and then $q$ implies $p$ by induction and that would satisfy it? I guess I have to do it twice? How would I go about proving something like this using induction? I'm confused what my base case would be, or even how to start. Any help would be greatly appreciated on this as right now I am very confused.

Answer 1

If you can show that $p_k \implies p_{k+1}$ and $p_n \implies p_{n-1}$, that should do it. Without knowing exactly what your problem is it is hard to provide help for your base case, but in essence you are using the technique twice, as you say. Maybe a base case to start could be $p_1 \implies p_2$?

Edit: As J.G. points out in a comment, another option is to show $p_1 \implies p_2 \implies \ldots \implies p_n \implies p_1$.

Answer 2

I have since realized that this is not the correct question that I needed to ask. I wasn't completely understanding the scenario when I proposed this question. I apologize for that. When I use the logical expression p1<-->p2<-->p3 for a 2^n truth table I've realized that when n is odd, the outcome of the truth table is only true when m is odd (m being the number of Pi that are true). Also when n is even, the outcome of the truth table is only true when m is even. That is what I supposed to realize after research. Now I need to figure out how to prove my formula using induction, and have no idea how to even set that up.