Let $$\begin{align} A1&=(p\implies (q\implies p)) \\ A2&=(((p\implies (q \implies r)) \implies ((p\implies q)\implies (p\implies r))) \\ A3&=((\neg p \implies \neg q ) \implies ((\neg p \implies q ) \implies p)) \end{align} $$
Let $S=(\mathcal{L}_{\{\neg,\implies,\wedge,\vee\}},\{A1,A2,A3\},\text{MP})$ be our system.
$\bf Theorem$
If $\Gamma \vdash \gamma $ then $\Gamma \vDash \gamma$.
Proof.
I'll do this by induction on the length $n$ of the proof for a formula $\tau$.
Base case. $n=1$.
Then the proof is
$$1.\tau\quad \text{Hypothesis.}$$
or $$1.\tau\quad \text{Axiom.}$$
In the first case, we have that $\tau\in \Gamma$, thus trivially, $\Gamma \vDash \tau$. In the second case, $\tau$ is a tautology, again, obviously, $\Gamma \vdash\tau$.
Now, let $\tau$ be a formula with a proof of length $n+1$.
Then that line looks like
$$n+1.\, \tau\quad \text{Axiom}$$
or
$$n+1.\, \tau\quad \text{MP (i,j).}$$
Where $i<j\leq n$.
On the first case, $\tau$ is a tautology, in the second case, we have formulas $\alpha=(\beta \implies \tau), \beta$ with proofs of length $i,j\leq n$ respectively, so we have that $\Gamma \vDash \alpha,\beta$.
Let $v$ be a valuation such that $v(\Gamma) = \{1\}.$ We have that $v(\beta \implies \tau)= 1$ if and only if $v(\beta)=0$ or $ v(\tau)=1$, but $\Gamma \vDash \beta$, therefore $v(\tau)=1$, and $\Gamma \vDash \tau$ Q.E.D.
Have I covered all the cases?
