How can the completeness of Hilbert's axioms be proven? How can one prove that every propositional tautology, expressed with

the connectives 
'$\neg$' and '$\rightarrow$', can be proved with the axioms below?
(P0. $\phi \to \phi$)
P1. $\phi \to \left( \psi \to \phi \right)$
P2. $\left( \phi \to \left( \psi \rightarrow \xi \right) \right) \to \left( \left( \phi \to \psi \right) \to  \left( \phi \to \xi \right) \right)$
P3. $\left ( \lnot \phi \to \lnot \psi \right) \to \left( \psi \to \phi \right)$
I'm especially interested in an eventual math-style proof:
Since all logical expressions have equivalents in form of elements in a Boolean ring with respect to XOR, AND and TRUE, and any tautology reduces to 1 in that ring, the Hilbert axioms can prove every tautology if they can prove all the axioms for a Boolean ring for the equivalents of $(1,\oplus,\cdot)$ expressed in only $(\neg,\rightarrow)$.
S1. $1\leftrightarrow (A\rightarrow A)$
S2. $AB\leftrightarrow\neg(A\rightarrow \neg B)$
S3. $A+B\leftrightarrow((A\rightarrow B)\rightarrow\neg(\neg A\rightarrow\neg B)) $
How to use P1-P3 to prove axioms of Boolean rings expressed with the substitution rules S1-S3? For example:


*

*the law of commutativity for multiplication: $\neg(A\rightarrow\neg
   B)\rightarrow\neg(B\rightarrow\neg A)$

*the law of multiplicative idempotence: $\neg(A\rightarrow\neg A)\rightarrow A$ and $A\rightarrow\neg(A\rightarrow\neg A)$

 A: Essentially you are asking why every tautology is provable, I post here a demonstration taken from my own notes.
$\varphi$ is any propositional fromula.
Let $v$ be a truth assignment and let $A_1, \ldots ,A_k$ 
be the propositional variables in $\varphi$.
Define 
$$\psi_i=\begin{cases} A_i & v(A_i)=T\\
\neg A_i & v(A_i)=F\end{cases}$$
And let 
$$\theta=\begin{cases} \varphi & v( \varphi)=T\\
\neg  \varphi & v( \varphi)=F\end{cases}$$
Then we claim
$$\{ \psi_1, \ldots , \psi_k \} \vdash \theta$$
This is proved by induction on the number of connectives in 
$\varphi$.
If there are no connectives then $\varphi$ is a 
propositional variable and the statement is trivial.
Assume $\varphi =\neg \eta$.  And let $v(\eta)=F$, 
then we have by induction 
$$\{ \psi_1, \ldots , \psi_k \} \vdash \neg \eta$$
which is the same as 
$$\{ \psi_1, \ldots , \psi_k \} \vdash \varphi$$ which is what is desired since 
$v(\varphi)=T$.
Now let $v(\eta)=T$
Then by induction
$$\{ \psi_1, \ldots , \psi_k \} \vdash  \eta.$$
In this case $v(\varphi)=F$, and so we have to show 
$$\{ \psi_1, \ldots , \psi_k \} \vdash \neg \varphi$$
or 
$$\{ \psi_1, \ldots , \psi_k \} \vdash \neg \neg \eta$$
But this follows since 
$\vdash \eta \rightarrow \neg \neg \eta$.
Now assume that 
$\varphi =\eta \rightarrow \sigma$
If $v(\varphi)=T$ then either $v(\sigma)=T$ in which case,
$$\{ \psi_1, \ldots , \psi_k \} \vdash \sigma $$
 and 
$$\vdash \sigma \rightarrow (\eta \rightarrow \sigma)$$ 
or 
$v(\eta)=F$ for which 
$$\{ \psi_1, \ldots , \psi_k \} \vdash \neg \eta $$
and 
$$\vdash \neg \eta \rightarrow (\eta \rightarrow \sigma).$$
The final case is where 
$v(\varphi)=F$ and so $v(\eta)=T$ and $v(\sigma)=F$.
Thus we have 
$$\{ \psi_1, \ldots , \psi_k \} \vdash \eta $$
$$\{ \psi_1, \ldots , \psi_k \} \vdash \neg \sigma $$
and from the last theorem we have 
$$\vdash \eta \rightarrow [\neg \sigma \rightarrow 
\neg ( \eta \rightarrow \sigma)].$$
Now to prove the completeness theorem, let $v$ and $w$ 
be truth assignments such that 
$v(A_i)=w(A_i)$ for $i<k$ and $v(A_k)\neq w(A_k)$
this means that if  $\varphi$ is a tautology,
we have 
$$\{ \psi_1, \ldots , \psi_{k-1}, A_k \} 
\vdash \varphi$$
and 
$$\{ \psi_1, \ldots , \psi_{k-1}, \neg A_k \} 
\vdash \varphi$$
therefore, 
$$\{ \psi_1, \ldots , \psi_{k-1}, \} 
\vdash \varphi$$
proceeding in this way we obtain 
$$\vdash \varphi.$$
A: See Derek Goldrei, Propositional and Predicate Calculus : A Model of Argument (2005).
See page 87 for the defnition of The formal system S that is based on your P2-P3-P4, of course with modus ponens.
See page 92 for the derivation of your P1 from P2 and P3.
See page 106 for the proof of 


Theorem 3.9 Completeness theorem for S.


