$\newcommand{\u}{{\uparrow}}\newcommand{\d}{{\downarrow}}$This question has been running through my mind a bit, so here's an attempt to implement Bram28's suggestion. I have not succeeded in giving a semantics for this language that simultaneously shows that none of the three formulas are derivable from the axioms, despite a fair number of tries. I also haven't quite pinned down where the difficulty comes from... perhaps someone else can add their answer in this regard. However, it becomes much easier if we regard the first formula as separate from the second and third.
Let us start with the (relatively) easy: formula 2 and 3. In this case we can take the domain $\{\top, \bot\}$ just like in conventional logic. In this interpretation of our logic, the truth value of a formula is inductively defined as
$$
\begin{align*}
[\u \phi] &= \top\\
[\d \phi] &= \bot\\
[\phi \to \psi] &= [\psi].
\end{align*}
$$
You can easily check that your axioms are satisfied; anything starting with an $\u$ is satisfied, and both the implications resolve to formulas starting with an $\u$ as well. Furthermore, note that modus ponens holds: if $[\phi] = \top$, and $[\phi \to \psi] = \top$, then trivially $[\psi] = \top$ -- that is just the second assumption. However, formulas two and three are not tautologies -- in fact, both evaluate to $\bot$, always.
This leaves the formula $\u\d A$. This one is a bit harder. We use the domain $\{0, 1, 2\}$, where 2 represents "truth", i.e. the axioms should evaluate to 2, the formula $\u\d A$ should not (always) evaluate to 2, and if both $[\phi]$ and $[\phi \to \psi]$ equal 2, then so should $[\psi]$. We choose the following interpretation:
$$
\begin{align*}
[\u \phi] &= \min([\phi] + 1, 2)\\
[\d \phi] &= 0\\
[\phi \to \psi] &= \begin{cases}
2 &\text{if $[\phi] \leq [\psi]$,}\\
0 &\text{otherwise.}
\end{cases}
\end{align*}
$$
Now we once again check by hand: the first axiom is satisfied, as any formula that starts with $\u\u$ is a tautology (note that actually you can prove this from the axioms, so this is a necessity). The second axiom is satisfied: $[\u\d\phi] = 1$, and $[\u\psi] \geq 1$, so $[\u\d\phi\to\u\psi] = 2$. Then consider the third axiom: we have $[\phi \to \psi] \in \{0, 2\}$; if it equals zero, the proposition will necessarily evaluate to $2$, so let us consider the case when $[\phi \to \psi] = 2$. Then $[\phi] \leq [\psi]$ by definition, and therefore also $[\u\phi] \leq [\u\psi]$ as the interpretation of $\u$ is monotone. Hence $[\u\phi \to \u\psi] = 2$, and the third axiom also always evaluates to 2.
Finally, modus ponens holds: if $[\phi]$ evaluates to 2, and $[\phi \to \psi]$ evaluates to 2, then $[\phi] \leq [\psi]$ and hence $[\psi]$ evaluates to 2.
However, the first formula in the list never evaluates to 2: in fact, it always evaluates to 1.
Note that in the first interpretation, the first formula does hold; and in the second interpretation, the final formula does not hold. Purely based on experimentation I feel that the two approaches are not commensurable, and would not be surprised if the smallest domain that shows all three formulas underivable has at least 4 elements.
