If something is proved for one set of axioms then you can use it for that set of axioms but wouldn't you have to prove it for another set of axioms before you could use if for the second set of axioms. So for instance must you prove 2+2=4 for every set of axioms before you can use it in any set of axioms, and is this for any piece of maths? P.S. Sorry about the title, could not think of an appropriate title, so feel free to change it or ask me to change it, thanks.
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The following answer is under the working assumption that we are talking about classical first-order logic.
Suppose that we can prove $\varphi$ from $T$. If $S$ is a theory such that $S\vdash T$, that is every axiom in $t$ is provable from $S$, then $\varphi$ is provable from $S$.
For example, consider the language including $0,s$ and $+$, where $s$ is an unary function whose meaning is "$sx$ is the successor of $x$". We don't have symbols for $1$ and $2$, but we write $1$ as a shorthand for $s0$ and $2$ as a shorthand for $ss0$.
Now we write the axioms: 1. $\forall x(x+0=x)$, and 2. $\forall x\forall y(x+s(y)=s(x+y))$.
We can now write a proof that $1+1=2$:
Now whenever we have $T$ whose language includes $s,0,+$ and $T$ proves the two axioms, we know that $T$ proves that $1+1=2$. This goes even further. If we can define (for every model of $T$) a constant $0$, a function $s$ and an operation $+$, and $T$ can prove the two axioms hold for those definable objects, then $T$ proves that $1+1=2$ for those as well.
On the other hand, in some cases we want to take something much weaker than $T$, and see if it still suffices to prove something. In those cases we need to actually work out the proof, or show that the two axioms hold for the addition. Or we can talk about a theory which is not weaker or stronger, but it is not provable from $T$ nor vice versa. In that case, again, we need to give some proof - and it might be a different one, depending on theory and the language.