$\mathbf{def}:$ a property $p^{op}$ is a dual property to $p$ if for all categories $\mathcal{C}$, $\mathcal{C}$ has $p^{op}$ iff $\mathcal{C}^{op}$ has $p$
$\mathbf{question 1}:$ is the above def is equivalent to the statement: a property $p^{op}$ is a dual property to $p$ if for all categories $\mathcal{C}$, $\mathcal{C}$ has $p$ iff $\mathcal{C}^{op}$ has $p^{op}$ . And why ?
$\bullet$ Let $ \Pi= \{q_{i} |i\in I\} $ be a set of properties and let $ \Pi^{op}= \{q_{i}^{op} |i\in I\} $ be the set of dual properties. Let $p$ be a single property. Consider the statement
$1)$ If a category $\mathcal{C}$ has $\Pi$, then it also has $p$.
Since all categories have the form $\mathcal{C}^{op}$ for some category $\mathcal{C}$, this statement is logically equivalent to the statement
$2)$ If a category $\mathcal{C}^{op}$ has $\Pi$, then it also has $p$.
and this is logically equivalent to
$3)$If a category $\mathcal{C}$ has $\Pi^{op}$ , then it also has $p^{op}$
$\mathbf{question 2}:$ Why $2$ and $3$ are equivalent.please explain with as much detail as possible. Thanks
Question 1 : yes : in the definition you have a universal quantification on categories so you can apply the definition to $C^{op}$; but also every category is the opposite of some category, because $(C^{op})^{op}= C$.
So whenever you have a statement that's universally quantified on $C$,you can change all $C$'s to $C^{op}$'s and all $C^{op}$'s to $C$'s and the statement will remain true.
Question 2 : 2 and 3 are equivalent by definition of dual property !
Assume 2 : let $C$ be a category with $\Pi^{op}$. Then $C^{op}$ has $\Pi$ by definition of dual property. Thus $C^{op}$ has $p$ by 2. Thus $C$ has $p^{op}$ by definition of dual property.
It's the same reasoning that shows that 3 => 2.