Is this the only solution to the problem? Find all mutually non-equivalent $A(p,q)$ so that $(Aq\rightarrow \overline{p}) = (p+A)$ is true regardless of $p$ and $q$, where $\overline{x}$ is negation, $xy$ is conjunction, $x+y$ disjunction, $x\rightarrow y$ implication and $x=y$ equivalence.
\begin{matrix}p&q&A&(Aq\rightarrow \overline{p}) = (p+A)\\
0&0&0&0\\
0&0&1&1\\
0&1&0&0\\
0&1&1&1\\
1&0&0&1\\
1&0&1&1\\
1&1&0&1\\
1&1&1&0
\end{matrix}
From the truth table we see that: $((Aq\rightarrow \overline{p}) = (p+A)) = \overline{p}\overline{q}A + \overline{p}qA + p\overline{q}\overline{A} + p\overline{q}A + pq\overline{A}$
$$\overline{p}\overline{q}A + \overline{p}qA + p\overline{q} + pq\overline{A} = 1$$
\begin{align}
A&=\overline{p}\overline{q}+\overline{p}q+\overline{(pq)}=\\
 &=\overline{p}+\overline{p}+\overline{q}=\\
 &=\overline{p}+\overline{q}
\end{align}
Cheking result
$$((\overline{p}+\overline{q})q\rightarrow \overline{p})=p+(\overline{p}+\overline{q})$$
For $p=0$ and $q=0$
$$
((1+1)0\rightarrow 1)=0+(1+1)\\
(0\rightarrow 1) = 1 \\
1=1 \\
$$
For $p=0$, $q=1$
$$
((1+0)1\rightarrow 1)=0+(1+0)\\
(1\rightarrow 1)=1 \\
1=1 \\
$$
For $p=1$, $q=0$
$$
((0+1)0\rightarrow 0)=1+(0+1)\\
(0\rightarrow 0)=1 \\
1=1 \\
$$
For $p=1$, $q=1$
$$
((0+0)1\rightarrow 0)=1+(0+0)\\
(0\rightarrow 0)=1 \\
1=1 \\
$$
This is a correct solution, but is it the only one?
I ask because the question uses the word all, and though that may not mean anything, I'd just like to check in case I missed something.
 A: It seems to me you are correct up to here:
$$\overline{p}\overline{q}A + \overline{p}qA + p\overline{q} + pq\overline{A}
 = 1.$$
It does not follow that
$A=\overline{p}\overline{q}+\overline{p}q+\overline{(pq)} 
= \overline p + \overline q,$ however.
Yes, it is true that
$$
(A = \overline p + \overline q)\rightarrow 
(\overline{p}\overline{q}A + \overline{p}qA + p\overline{q} + pq\overline{A}
 = 1)
$$
for all possible truth assignments,
but the converse is not true.
The truth table shows that $A$ is one of two possible non-equivalent expressions, because for just one assignment to $p$ and $q,$
namely, $p=1, q=0,$ you have two choices for the value of $A.$

In particular, try the substitution $A = \overline p.$
With that substitution,
$Aq\rightarrow \overline p$ becomes
$\overline pq\rightarrow \overline p$, a tautology,
and $p+A$ becomes
$p+\overline p$, also a tautology,
so $(Aq\rightarrow \overline p)=(p+A).$
You can find this solution (rather than having to guess it) by selecting rows $2$, $4$, $5$, and $7$ from the truth table.
A: Line 5 of the table shows that while  $Aq\rightarrow \overline{p} = p+A$ is satisfied, you do not have that $A = \overline{p}+\overline{q}$, and so it is not even a solution to the equation.
Moreover, lines 5 and 6 demonstrate that you cannot express $A$ as a function of $p$ and $q$ to get all solutions, since in both lines $Aq\rightarrow \overline{p} = p+A$ is satisfied, and while the values of $p$ and $q$ are the same, the value of $A$ is not the same.
A: Just a tip for you, never write im terrbily sorry bc my english is so bad,because that tells me and others either that you are looking for compliments or you actually made grammar mistakes which makes the situation even worse because now we focus on the language rather than the mathematics. I know its polite to acknowledge weaknesses in some cultures but in stack exchange culture it is just a waste of alpah-betas
