Double Induction Principle/Lemma
Lemma for Double Induction Principle
Let $M$ be a minimally inductive class under a mapping $g$.
Let $\RR$ be a relation on $M$ which satisfies:
\((\text D_1)\) | $:$ | \(\ds \forall x \in M:\) | \(\ds \map \RR {x, \O} \) | ||||||
\((\text D_2)\) | $:$ | \(\ds \forall x, y \in M:\) | \(\ds \map \RR {x, y} \land \map \RR {y, x} \implies \map \RR {x, \map g y} \) |
Let $x$ be a right normal element of $M$ with respect to $\RR$.
Then $x$ is also a left normal element of $M$ with respect to $\RR$.
Proof
The proof proceeds by general induction.
Let $x \in M$ be right normal with respect to $\RR$.
Let $\map P y$ be the proposition:
- $\map \RR {x, y}$ holds.
Basis for the Induction
By condition $\text D_1$ of the definition of $\RR$:
- $\map \RR {x, \O}$
for all $x \in M$.
Thus $\map P \O$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that if $\map P y$ is true, where $x \in M$, then it logically follows that $\map P {\map g y}$ is true.
So this is the induction hypothesis:
- $\map \RR {x, y}$ holds
from which it is to be shown that:
- $\map \RR {x, \map g y}$ holds.
Induction Step
This is the induction step:
Let $\map \RR {x, y}$ hold.
As $x$ is right normal with respect to $\RR$:
- $\map \RR {y, x}$ holds.
Thus by condition $\text D_2$ of the definition of $\RR$:
- $\map \RR {x, \map g y}$ holds.
So $\map P x \implies \map P {\map g x}$ and the result follows by the Principle of General Induction.
Therefore:
- $\forall x \in M$: if $x$ is a right normal element of $M$ with respect to $\RR$, then $x$ is a left normal element of $M$ with respect to $\RR$
$\blacksquare$
Sources
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Theorem $4.3$