How can I prove, preferably using first-order logic, that if the graphs of 2 functions are equal, then the domains of the 2 functions are also equal?

Question

I know that, by definition, two functions are equal iff, among other things, their domains are equal. This is why I specified that the two functions only have the same graph, but are not neccesarily equal.

Answer 1

The graph of a function is, in set-theoretic terms, the function itself: $f= \{ (x,y) \mid x∈A \text { and } y∈B \}$.

And the domain of $f$ is: $\text {Dom}(f)= \{ x \mid ∃y((x,y) ∈ f) \}$.

We have that $f=g$ means that $(x,y) \in f ↔ (x,y) \in g$.

Thus, the proof is:

$x ∈ \text {Dom}(f) ↔ ∃y((x,y) ∈f) ↔ ∃y((x,y) ∈ g) ↔ x ∈ \text {Dom}(g)$.

Answer 2

$f$ & $g$ have a common graph implies that for each $x\in\mathbb R,$ either $f(x){=}g(x)$ or $f(x)$ & $g(x)$ are both undefined, which implies that for each $x\in\mathbb R,$ either $(x\in\operatorname{Dom}_f$ & $x\in\operatorname{Dom}_g)$ or $(x\notin\operatorname{Dom}_f$ & $x\notin\operatorname{Dom}_g),$ which implies that $f$ & $g$ have a common domain.