ThmDex – An index of mathematical definitions, results, and conjectures.
Proof by principle of weak mathematical induction

Let $P = (\mathbb{Z}, {\leq})$ be the D1098: Ordered set of integers such that
 (i) $f : \mathbb{Z} \to \{ 0, 1 \}$ is a D218: Boolean function on $\mathbb{Z}$ (ii) $a \in \mathbb{Z}$ is an D995: Integer (iii) $$[a, \infty) : = \{ n \in \mathbb{Z} : n \geq a \}$$ (iv) $$X : = \{ n \in \mathbb{Z} : f(n) = 1 \}$$ (v) $$a \in X \subseteq [a, \infty)$$ (vi) $$\forall \, n \in \mathbb{Z} \left( n \in X \quad \implies \quad n + 1 \in X \right)$$
Then $$X = [a, \infty)$$
Remarks
 ▶ Remark 0 In a proof by weak induction, we interpret $f$ to be a predicate statement depending on an integer $n \in \mathbb{Z}$ which we want to show to be true for all integers on some ray $[a, \infty) : = \{ n \in \mathbb{Z} : n \geq a \}$ where the statement being true for $n$ is understood to be encoded by $f(n) = 1$. To accomplish this, we must show that $\{ n \in \mathbb{Z} : f(n) = 1\} = [a, \infty)$.
Proofs
Proof 0
Let $P = (\mathbb{Z}, {\leq})$ be the D1098: Ordered set of integers such that
 (i) $f : \mathbb{Z} \to \{ 0, 1 \}$ is a D218: Boolean function on $\mathbb{Z}$ (ii) $a \in \mathbb{Z}$ is an D995: Integer (iii) $$[a, \infty) : = \{ n \in \mathbb{Z} : n \geq a \}$$ (iv) $$X : = \{ n \in \mathbb{Z} : f(n) = 1 \}$$ (v) $$a \in X \subseteq [a, \infty)$$ (vi) $$\forall \, n \in \mathbb{Z} \left( n \in X \quad \implies \quad n + 1 \in X \right)$$
This result is a particular case of R796: Principle of weak mathematical induction. $\square$