Fractional derivative of fractional integral equal identity

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Consider the definition of Riemann-Liouville fractional order derivative as follows $$^{RL}_a\mathcal{D}^\alpha_xf(x)=\frac{1}{\Gamma(1-\alpha)}\frac{d}{dx}\int_a^x \frac{f(u)}{(x-u)^{\alpha}}du,\qquad x\geq a,\quad 0<\alpha<1$$and the fractional order integral: $$^{RL}_a\mathcal{I}^\alpha_xf(x)=^{RL}_a\mathcal{D}^{-\alpha}_xf(x)=\frac{1}{\Gamma(\alpha)}\int_a^x \frac{f(v)}{(x-v)^{1-\alpha}}dv,\qquad x\geq a,\quad 0<\alpha<1.$$ Then $$^{RL}_a\mathcal{D}^\alpha_x\;(_a\mathcal{I}^\alpha_xf(x))=f(x)\qquad (1)$$

and $$^{RL}_a\mathcal{I}^\alpha_x\;(_a\mathcal{D}^\alpha_xf(x))\neq f(x)\qquad (2) $$ I'm wondering if someone knows the analytical proof of the equation (1) ?

One can found the two results and their proof in the following ref:

$\small{ \text{Gorenflo, Rudolf, and Francesco Mainardi. "Fractional calculus:}}$ $\small{\text{integral and differential equations of fractional order." arXiv preprint arXiv:0805.3823 (2008).}}$

https://arxiv.org/pdf/0805.3823.pdf