Crossed induction

Question

I have two sequences of equalities $L_n$ and $R_n$.

The equation $L_{n+1}$ is true only if $R_n$ is true, and the same happens for $R_{n+1}$

$$R_n \implies L_{n+1}$$

$$L_n \implies R_{n+1}$$

How can I use induction in order to prove the sequences $L_n$ and $R_n$.

My original idea was to prove $R_0$ and $L_0$, then show the proof of $R_n \implies L_{n+1}$ and $L_n \implies R_{n+1}$. But I'm not sure that this is enough... I feel like there is something I'm missing.

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I wonder if this could work for more sequences like $A_n$, $B_n$ and $C_n$ with

$A_n \implies B_{n+1}$

$B_n \implies C_{n+1}$

$C_n \implies A_{n+1}$

Answer 1

It is quite evident that $R_n\wedge L_n\Rightarrow R_{n+1}\wedge L_{n+1}$ so induction can be applied here.

Answer 2

I think it is correct, for the initial cases

$R_0$ is true and $L_0 ⟹ R_1$ is true

and the induction process

$$R_n ⟹ L_{n+1} ⟹ R_{n+2}$$

we can conclude $R_n$ is true for all $n \geq 0$. Symmetric for $L_n$.

Answer 3

$R_1\Rightarrow L_2\Rightarrow R_3\Rightarrow L_4\Rightarrow\dots$

and

$L_1\Rightarrow R_2\Rightarrow L_3\Rightarrow R_4\Rightarrow\dots$

So once you've established the base cases of $R_1$ and $L_1$, the rest follows.

A similar thing would happen for your question about $A_n, B_n$, and $C_n$.