The base scheme is an algebraically closed field.

Let $X\to \mathbf{P}^1$ be an arithmetic surface over $\mathbf{P}^1$ and let $P$ be a section of $X\to \mathbf{P}^1$. Let $D$ be an effective (*edited*) divisor on $X$. Is it trivial that the intersection product $D\cdot P \geq \mathrm{ord}_P(D) (P\cdot P)$? Here $\mathrm{ord}_P(D)$ is the order of $D$ at $P$, i.e., the coefficient of $D$ at $P$.

Here's my idea of a proof.

If $E$ is an integral divisor on $X$ and the image of $P$ is not contained in the support of $E$, then $E\cdot P \geq 0$. Therefore, we have that $D\cdot P \geq \mathrm{ord}_P(D) (P\cdot P)$.

Can we replace $P$ by any integral horizontal divisor? Can we replace $\mathbf{P}^1$ by a Dedekind scheme?

*Remark.* An arithmetic surface over $\mathbf{P}^1$ is a flat projective morphism $X\to \mathbf{P}^1$ with $X$ an integral regular and $2$-dimensional scheme.