I wondered about this and I am having a hard time formulating it as a question at all, but I hope I can express something if my wondering here.
Assume we have a set $V = \{\mathbb N, +, =, X ,(,)\}$. Here $\mathbb N$ is the set of all natural numbers. The set $V$ contains our alphabet.
We also have set $A = \{ "1=1", "1+1=1+1", \dots, "1+1=2","1+1+1=3", \dots \}$. Set $A$ contains all the axioms. These are actually more like definitions, I believe, but they are axioms too.
Then let the set $T$ contains as elements all true statements. We know $A \subset T.$
Finally define binary operator $X$ as a function from $T^2 \to T$ such that if $a,b \in T, \ aXb \in T$, where $b$ is "applied" is some way (which seems rather difficult to define) to $a$.
A statement $s$ made using the elements of $V$ is in $T$ if $s = a_1Xa_2Xa_3...$ for some set of axioms, ie. $\forall i, a_i \in A$.
Assume we want to know if "theorem" $2+2 = 4$ is an element of $T$. Let's first name some axioms for simplicity: $$"1+1+1+1=1+1+1+1" = a_1 \\ "2 = 1+1" = a_2 \\ "4 = 1+1+1+1" = a_3$$
We can see, assuming a liberal definition of $X$: $$"1+1+1+1=1+1+1+1" X "2 = 1+1" = "2+1+1=1+1+1+1"$$ $$"2+1+1=1+1+1+1" X "2 = 1+1" = "2+2=1+1+1+1"$$ $$"2+2=1+1+1+1" X "4 = 1+1+1+1" = "2+2=4".$$ Thus $$"2+2 = 4" = (((a_1Xa_2)Xa_2)Xa_3).$$ Thus 2+2=4 is a true statement and thus an element of $T$.
Some questions: Does this make sense? Can such a binary operator on axioms be defined? Could it be applied to "real" sets of axioms? Could the set of true statements form a group under such operator?