Can Hilbert (style) prove my tautology? Can Hilbert (style) axioms prove the following tautology?

$$A\wedge(C\rightarrow B)\oplus(A\wedge B\leftrightarrow C)\rightarrow(C\rightarrow(A\rightarrow B))\qquad\text{algebraic style}
$$$$
(((A\wedge(C\rightarrow B))\oplus((A\wedge B)\leftrightarrow C)))\rightarrow(C\rightarrow(A\rightarrow B))\qquad\text{formal style}
$$

Here $\wedge$ means logical and, $\oplus$ means exclusive-or, and $\leftrightarrow$ means logical equivalence.

Due to Wikipedia/Hilbert style the axioms below describe classical propositional logic:

P1. $\phi \to \phi$

P2. $\phi \to \left( \psi \to \phi \right)$

P3. $\left( \phi \to \left( \psi \rightarrow \xi \right) \right) \to \left( \left( \phi \to \psi \right) \to  \left( \phi \to \xi \right) \right)$

P4. $\left ( \lnot \phi \to \lnot \psi \right) \to \left( \psi \to \phi \right)$

How to go? Is it necessary to add definitions of connectives?
You can see Elliott Mendelson, Introduction to mathematical logic (4ed - 1997) : PROPOSITION 1.14 (COMPLETENESS THEOREM).

The proof (due to Kalmar,1935) allows us to build a proof of a tautology $\mathcal B$ whatever.

Assume $\mathcal B$ is a tautology, and let $B_1, \ldots, B_k$ be the statement letetrs in $\mathcal B$. For any truth value assignment to $B_1, \ldots, B_k$ we have : 


  $B_1', \ldots, B_k' \vdash \mathcal B$


where let $B_i'$ be $B_i$ if $B_i$ takes the value $T$, and let $B_i'$ be $\lnot B_i$ if $B_i$ takes the value $F$.

Thus, due to the fact that your formula $\mathcal F$ has only three statement letters : $A,B,C$, it is enough to rewrite it in "primitive" notation, i.e. using only $\rightarrow$ and $\lnot$ (call it : $\mathcal F'$) and then apply the procedure described in the above Proposition.

Starting from example from the truth value assignment $v(A)=v(B)=v(C)=T$, we have :


  
    $A,B,C \vdash \mathcal F'$.
To do this you would want to first define the other connectives.  I will use Lukasiewicz/Polish notation.  For your "∧" I write "K".  For your "→" I write "C".  For your "⊕", following A. N. Prior, I will write "J".  For "$\lnot$", "N".  For "↔" I will write E.

The axioms become 

P2 CpCqp.

P3 CCpCqrCCpqCpr.

P4 CCNpNqCqp.  

I will note that P1 is NOT an axiom given that a requirement for axioms is that they are independent, and I simply fail to understand this new standard where axioms don't have to be independent.

Now we'll need definitions for the other connectives, which in effect will end up behaving like axioms.  The symbol "$\delta$" indicated a variable function of one argument.  This means we can make substitutions for $\delta$.

PK: C$\delta$Kpq$\delta$NCpNq.

PE: C$\delta$Epq$\delta$NCCpqNCqp.

PJ: C$\delta$Jpq$\delta$CCpqNCqp.

Your formula is CJKaCcbEKabcCcCab.

Using definition PK we can tell that  CJKaCcbEKabcCcCab is equivalent to

CJNCaNCcbENCaNbcCcCab.

Using definition PE we can tell that CJNCaNCcbENCaNbcCcCab is equivalent to

CJNCaNCcbNCCNCaNbcNCcNCaNbCcCab.

Using definiton PJ we can tell that CJNCaNCcbNCCNCaNbcNCcNCaNbCcCab is equivalent to

CCCNCaNCcbNCCNCaNbcNCcNCaNbNCNCCNCaNbcNCcNCaNbNCaNCcbCcCab.

To try to prove it using Prover9, or OTTER you'll only need to parenthesize all of the axioms and use a unary predicate (which you may think of as meaning "provable" or "is formally provable"), and rewrite the variables as "x", "y" and "z" respectively for "a", "c", and "b" or "p", "q" and "r"... something like this:

P2: P(C(x,C(y,x))).

P3: P(C(C(x,C(y,z)),C(C(x,y),C(x,z)))).

P4: P(C(C(N(x),N(y)),C(y,x))).

Parenthesize P(CCCNCxNCzyNCCNCxNyzNCzNCxNyNCNCCNCxNyzNCzNCxNyNCxNCzyCzCxy), and add this as the goal.  This formula is:

 P(C(C(C(N(C(x,N(C(z,y)))),N(C(C(N(C(x,N(y))),z),N(C(z,N(C(x,N(y)))))))),N(C(N(C(C(N(C(x,N(y))),z),N(C(z,N(C(x,N(y))))))),N(C(x,N(C(z,y))))))),C(z,C(x,y)))).


You'll also want an assumption which enables Prover9 to prove things like it were using condensed detachment:

-P(C(x,y)) | -P(x) | P(y).  (it is not the case that either it is provable that x implies y, or it is not the case that x is provable, or y is provable).  Then you might find a proof equivalent to a proof using condensed detachment.  You'll want to prove that since you have

PK: C$\delta$Kpq$\delta$NCpNq.

PE: C$\delta$Epq$\delta$NCCpqNCqp.

PJ: C$\delta$Jpq$\delta$CCpqNCqp.

you can prove 

PK reversed: C $\delta$NCpNq $\delta$Kpq.

PE reversed: C $\delta$NCCpqNCqp $\delta$Epq.

PJ reversed: C $\delta$CCpqNCqp $\delta$Jpq.

And it does come as possible to prove each of those, since it has gotten proven that from 

C $\delta$ P $\delta$ R 

we can derive

C $\delta$ R $\delta$ P

where "R" and "P" are constants, and we only have substitution and detachment as the rules of inference.

Another approach could involve proving CCNpqCCNpNqp (or CCNpNqCCNpqp).  Then one could set up a natural-deduction like demonstration of the C-N fomrula you've referred to, and then use the proof of the deduction meta-theorem to produce a formal proof.