Proof that two antiderivatives differ by a constant, quantifier confusion

The proof that two antiderivatives of f differ by a constant is given in several other questions.

By the mean value theorem, for all $a<b\in I$, there exists an $x\in [a,b]$ such that

$(F-G{)}^{\mathrm{\prime}}(x)=\frac{(F-G)(b)-(F-G)(a)}{b-a}\phantom{\rule{0ex}{0ex}}={F}^{\mathrm{\prime}}(x)-{G}^{\mathrm{\prime}}(x)\phantom{\rule{0ex}{0ex}}=f(x)-f(x)\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}0=\frac{(F-G)(b)-(F-G)(a)}{b-a}\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(F-G)(b)=(F-G)(a)\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}F(b)-G(b)=F(a)-G(a)$

so we can define

$C:=F(b)-G(b)=F(a)-G(a)$

But since that's true for all a and b, what happens if we let $a=infI$? Then $F(a)=G(a)=0,$, which means that $F(b)=G(b)$ for all $b>infI$. But that doesn't seem like it's true.

So I suspect I'm confused about quantifiers somewhere. Where am I going wrong?