Second Principle of Mathematical Induction/Zero-Based

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Let $\map P n$ be a propositional function depending on $n \in \N$.

Suppose that:

$(1): \quad \map P 0$ is true
$(2): \quad \forall k \in \N: \map P 0 \land \map P 1 \land \ldots \land \map P {k - 1} \land \map P k \implies \map P {k + 1}$


$\map P n$ is true for all $n \in \N$.


For each $n \in \N$, let $\map {P'} n$ be defined as:

$\map {P'} n := \map P 0 \land \dots \land \map P n$

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

It is immediate from the assumption $\map P 0$ that $\map {P'} 0$ 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 $n \in \N$

as desired.


Also see

  • Results about Proofs by Induction can be found here.