compound proposition logically equivalent I can not solve this question 
Find a compound proposition logically equivalent to $p \to q$ using only the logical operator $\downarrow$.
 A: It's quite simple; let start with :

$\lnot p \equiv p \downarrow  p$.

Then :


$p \rightarrow q \equiv ((p \downarrow p) \downarrow q) \downarrow ((p \downarrow p) \downarrow q)$.


In order to check the definition, we have to use the truth-table for $\downarrow$ : it is true only when both $p$ and $q$ are false. 
This fact justify the definition of $\lnot p$ as $p \downarrow p$. 
For the conditional, we will work by steps: 

$((p \downarrow p) \downarrow q)$ is $\lnot p \downarrow q$ [see the definition of $\lnot$ above]. 

Thus, the complete formula is simply : $\lnot ( \lnot p \downarrow q)$.
Now we may check that the only case when it is false (i.e.$0$) is when $p=1$ and $q=0$. 
Note. You can see this paper on Adeqaute set of connectives for a general overview of the topic.
A: Given propositions $p$ and $q$, the propositon $p \downarrow q$ is given by the following truth table:
$$
\begin{align}
p \qquad &  q  & p \downarrow q \\
T \qquad & T  & F \\
T \qquad & F  & F \\
F \qquad & T  & F \\
F \qquad & F  & T 
\end{align} 
$$
Thus we have
$$
T \downarrow T \equiv F, \ \ \ T \downarrow F \equiv F, \ \ \ F \downarrow T \equiv F, \ \ \  F \downarrow F \equiv T. \tag{0} 
$$
In fact, we have the logical identity
$$
p \lor q \equiv \overline{p \downarrow q}. \tag{1} 
$$
We further note that
$$ p \downarrow p \equiv \overline{p}, \tag{2} $$
because when $p$ is True, we have $T \downarrow T$, which is False, and when $p$ is False, we have $F \downarrow F$, which is true.
And, we also note that
$$
\begin{align} 
p \lor q &\equiv \overline{\overline{p \lor q}} \\ &\equiv \overline{p \downarrow q} \qquad \mbox{[ using (1) above ]}\\
&\equiv \big( p \downarrow q \big) \downarrow \big( p \downarrow q \big).
\end{align}
$$
Thus we have the logical identity
$$
p \lor q \equiv \big( p \downarrow q \big) \downarrow \big( p \downarrow q \big). \tag{3} 
$$
Now we note that
$$
\begin{align}
p \rightarrow q &\equiv \overline{p} \lor q \qquad \mbox{[ using the conditional-disjunction equivalence ] } \\
&\equiv \left( \overline{p} \downarrow q \right) \downarrow \left( \overline{p} \downarrow q \right) \qquad \mbox{[ using (3) above with $\overline{p}$ in place of $p$ ]} \\
&\equiv \big( (p \downarrow p) \downarrow q \big) \downarrow \big( (p \downarrow p) \downarrow q \big) \qquad \mbox{[ using (2) above ]},
\end{align}
$$
as required.
