From this link
If $X$ is an $\infty$-groupoid, then a property $P$ of objects of $X$ is compatible with equivalence if, whenever $P(a)$ holds for an object $a$ of $X$ and $b$ is equivalent (as an object of $X$) to $a$, then $P(b)$ holds. Alternatively, an operation $f$ from objects of $X$ to objects of (another $\infty$-groupoid) $Y$ is compatible with equivalence if, whenever $a$ and $b$ are equivalent objects of $X$, $f(a)$ and $f(b)$ are equivalent (as objects of $Y$).
in the same link is written:
The operation-version of the principle of equivalence gives the property-version if you think of a property as an operation taking values in the $\infty$-groupoid (in fact a $0$-groupoid, or set) of truth values. (The property-version also gives the operation-version, although that is a little more involved.)
Here is my question: how can you recast the operation-version of the principle of equivalence in the property-version?
I think I finally got it.
For each operation $f$ defined between two (object-sets) of two $\infty$-groupoids $X$ and $Y$ and each object $a \in X$ we can define the property $$P_{f,a}(x) \equiv f(a) \sim f(x)$$ where $x \sim y$ stands for $x$ is equivalent to $y$.
Then the operation $f$ satisfies the principle of equivalence (i.e. is compatible with equivalence) if and only if for each $a \in X$ the property $P_{f,a}(x)$ is compatible with equivalence.
Indeed if $f$ is compatible with equivalence then for every $a \in X$ if $b,c \in X$ are such that $b \sim c$ then we have $$P_{f,a}(b) \equiv f(a) \sim f(b)\ ,$$ $$P_{f,a}(c) \equiv f(a) \sim f(c)$$ and $$f(b) \sim f(c)$$ hence, since $\sim$ is an equivalence relation, it follow that $P_{f,a}(b) \iff P_{f,a}(c)$.
On the other hand if for each $a \in X$ the property $P_{f,a}(x)$ satisfies the principle of equivalence then for each pair $a,b \in X$ if $a \sim b$ we have that $$P_{f,a}(a) \equiv f(a) \sim f(a)$$ is clearly true and so $$P_{f,a}(b) \equiv f(a) \sim f(b)$$ must be true too.