Logical question about the fundamental theorem of calculus

So if f is continuous on [a,b] and F is any antiderivative of f, and $I=[a,b]$ then ${\int}_{I}f=F(b)-F(a)$.

Okay So the main proof for this as follows :

Let ${F}_{0},{F}_{1}$ be particular antiderivatives, We have ${F}_{1}-{F}_{0}=C\text{}=\text{}{F}_{1}={F}_{0}+C$ let ${I}_{1}=[a,x]$ then since; ${F}_{0}={\int}_{{I}_{1}}f$.

We have ${F}_{0}(a)=0=>{F}_{1}(a)=C$

Then the proof is almost completed. But my problem is, if for any particular antiderivative ${F}_{0}(a)=0$ wouldn't ${F}_{1}(a)=0$ be the case, since what we did here was to use the definition of antiderivative, we could easily write ${F}_{1}={\int}_{{I}_{1}}f$

So if f is continuous on [a,b] and F is any antiderivative of f, and $I=[a,b]$ then ${\int}_{I}f=F(b)-F(a)$.

Okay So the main proof for this as follows :

Let ${F}_{0},{F}_{1}$ be particular antiderivatives, We have ${F}_{1}-{F}_{0}=C\text{}=\text{}{F}_{1}={F}_{0}+C$ let ${I}_{1}=[a,x]$ then since; ${F}_{0}={\int}_{{I}_{1}}f$.

We have ${F}_{0}(a)=0=>{F}_{1}(a)=C$

Then the proof is almost completed. But my problem is, if for any particular antiderivative ${F}_{0}(a)=0$ wouldn't ${F}_{1}(a)=0$ be the case, since what we did here was to use the definition of antiderivative, we could easily write ${F}_{1}={\int}_{{I}_{1}}f$