I've seen continuity in R defined in the two following ways. Namely
$$\forall c \in X, \forall \epsilon > 0 , \exists \delta>0, ((x\in X \land |c-x|<\delta) \implies |f(c)-f(x)| < \epsilon) $$
$$\forall c \in X, \forall \epsilon > 0 , \exists \delta>0, \forall x\in X (|c-x|<\delta \implies |f(c)-f(x)| < \epsilon) $$
Thus (letting $A = |c-x|<\delta$ and $ B= |f(c)-f(x)| < \epsilon$) it must be that for statements $A,B$, the following are equivalent.
$$\forall x \in X,(A\implies B) $$ $$(x \in X \land A) \implies B$$
I'd like to prove this formally, but I'm not sure how. I tried calling both $x\in X$ and $\forall x \in X$ statement $C$, and making a truth table. I found that $$C \land (A\implies B) \equiv (C \land A) \implies B$$ iff $C$ is true. I'm quite confused; do quantifiers like $\forall x \in X$ even have truth values?
No, quantifiers do not have truth values, on their own. However a restricted quantifier has equivalent forms, and we can use universal generalisation.
$$\begin{align}(\forall x\in X) (A(x)\,\to\, B(x)) ~&\iff (\forall x)\Big(x\in X\to\big(A(x)\to B(x)\big)\Big) \\[1ex] ~&\iff (\forall x)\big((x\in X\,\wedge\, A(x))\,\to\, B(x)\big)\\[1ex] &\;\implies (x_1\in X\,\wedge\, A(x_1))\;\to\; B(x_1)\end{align}$$
So the formal definition is: $~\forall c {\in} X~ \forall \varepsilon {>} 0 ~ \exists \delta{>}0 ~ \forall x{\in} X ~\big(\lvert c-x\rvert<\delta ~\to~ \lvert f(c)-f(x)\rvert < \varepsilon\big)~$ where all variables are bound.
Infering a generalised form: $\forall c {\in} X~ \forall \varepsilon {>} 0 ~ \exists \delta{>}0 ~ \big(\big(x\in X~\wedge~\lvert c-x\rvert<\delta\big) ~\to~ \lvert f(c)-f(x)\rvert < \varepsilon\big)$ where the arbitrary variable, $x$, is free.
Note:
The first form is the stronger uniform continuity, and states that there exists a $\delta$ for all $x$, which satisfies the predicate.
The second form is weaker continuity; it states that for each arbitrary value, $x$, there is some $\delta$ value that satisfies the predicate.