Polynomials of Congruent Ring Elements are Congruent

Theorem
Let $R$ be a commutative ring with unity.

Let $I$ be an ideal of $R$.

Let $x, y \in R$.

Let:
 * $x \equiv y \pmod I$

where the notation indicates congruence modulo $I$.

Let $\map F X \in R \sqbrk X$ be a polynomial in one variable over $R$.

Then:
 * $\ds \map F x \equiv \map F y \pmod I$

Proof
Let $\map F X = \ds \sum_{k \mathop = 0}^r a_k X^k$ where $X$ is the indeterminate and $a_0, a_1, \ldots, a_r \in R$.

It has to be shown:
 * $\ds \sum_{k \mathop = 0}^r a_k x^k \equiv \sum_{k \mathop = 0}^r a_k y^k \pmod I$

From Left Cosets are Equal iff Product with Inverse in Subgroup:
 * $\forall a, b \in R: a \equiv b \pmod I \iff a + I = b + I$

Now: