The modified Bessel function of the 1st kind $I_0$ is defined by $$ I_0(z)=\frac1\pi\int_0^{2\pi}e^{z\cos\theta}\,d\theta $$ and arises, among other places, in the probability density function of a noncentral $\chi^2(2)$ random variable.
The error function is defined by $$ \text{erf}(z)=\frac2{\sqrt\pi}\int_0^z e^{-t^2}dt $$ and is closely related to the cumulative distribution function for the standard normal distribution.
I believe these are not elementary functions, but
is either $I_0$ or $\text{erf}$ elementary "relative to" the other?
That is, if we were to add one of them to our repertoire of "elementary" functions, would the other one become "elementary"?
Edit: @Suvrit mentioned that with the confluent hypergeometric function $_1F_1=M$, $$ \text{erf}(z)=\frac2{\sqrt\pi} M\left(\frac12,\frac32,-z^2\right)\cdot z\qquad\text{and} $$ $$ I_0(z)= M\left(\frac12,1,2z\right)e^{-z},\qquad\text{where} $$ $$M(1/2;b;z)=\sum_{n=0}^\infty \frac{(1/2)^{(n)}z^n}{b^{(n)}n!}\qquad\text{and}$$ $$ a^{(n)}=a(a+1)\cdots (a+n-1) $$ so that $a^{(2)}=(a+1)a$ etc. In particular $$(1/2)^{(n)} = \frac12 \frac32 \frac52 \dots = \frac{(2n)!}{n!4^n}\qquad\text{and}$$ $$\frac{(1/2)^{(n)}}{(3/2)^{(n)}}=\frac{\frac12\frac32\frac52\dots\frac{2n-1}{2}}{\frac32\frac52\frac72\dots\frac{2n+1}{2}}=\frac{1/2}{(2n+1)/2}=\frac1{2n+1}.$$ So the question can be rephrased as:
are the following functions elementary relative to eachother? \begin{align} f(z):=M\left(\frac12,\frac32,-z^2\right)&=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}\frac{z^{2n}}{n!},&\qquad\text{and}\\ g(z):=M\left(\frac12,1,2z\right)&=\sum_{n=0}^\infty \frac{(2n)!}{n!2^n}\frac{z^n}{n!}.& \end{align}