How to prove Corollary Let $k \in \mathbb{Z}$, and $P(x)$ be a formula in one variable such that (1) $P(k)$ and (2) $(\forall x ≥k)P(x)\implies P(x+1$

Question

Prove Corollary

Let $k \in \mathbb{Z}$, and $P(x)$ be a formula in one variable such that

(1) $P(k)$

(2) $(\forall x ≥k)P(x)\implies P(x+1)$

Then

$$\forall \in \mathbb{Z}x≥k \implies P(x)$$

Question This can be proved by defining a new formula that can be proved with standard induction.Can you define the formula?

Assume I can define the new formula as $Q(k)$ and $k \in \mathbb{N}$

such that $\forall x ≥ k \ P(x)$

Then I could prove $Q(k)$ by induction

base case $Q(0)\implies P(0)$

Then assume $Q(k)\implies P(k+1),P(k+2),...,P(k+n),n\in \mathbb{N}$

And $Q(k+1) \implies P(k+2),...,p(k+n)$

Is this logic valid?