Second Principle of Mathematical Induction/One-Based

Theorem
Let $\map P n$ be a propositional function depending on $n \in \N_{>0}$.

Suppose that:


 * $(1): \quad \map P 1$ is true


 * $(2): \quad \forall k \in \N_{>0}: \map P 1 \land \map P 2 \land \ldots \land \map P {k - 1} \land \map P k \implies \map P {k + 1}$

Then:


 * $\map P n$ is true for all $n \in \N_{>0}$.

Proof
For each $n \in \N_{>0}$, let $\map {P'} n$ be defined as:


 * $\map {P'} n := \map P 1 \land \dots \land \map P n$

It suffices to show that $\map {P'} n$ is true for all $n \in \N_{>0}$.

It is immediate from the assumption $\map P 1$ that $\map {P'} 1$ is true.

Now suppose that $\map {P'} n$ holds.

By $(2)$, this implies that $\map P {n + 1}$ holds as well.

Consequently, $\map {P'} n \land \map P {n + 1} = \map {P'} {n + 1}$ holds.

Thus by the Principle of Mathematical Induction:


 * $\map {P'} n$ holds for all $n \in \N_{>0}$

as desired.