$\infty$-groupoids = Kan complexes

Question

One direction is clear, a Kan complex has all its morphisms invertible hence it is an $\infty$-groupoid. Is it possible to show the opposite as a Corollary of the following theorem?
Theorem. Let $p:X\to Y$ an isofibration between $\infty$-categories. For any monomorphism $i:A\hookrightarrow B$ in sSet, the canonical induced map $$K(B,X)\to K(A,X)\times_{K(A,Y)} K(B,Y)$$ is a Kan fibration between Kan complexes.
Here $K(B,X)=\{f:\Delta^n\times B\to X| \forall b\in B,f_b\in X^\simeq_n\}$
As a special case, one can see that any $K(B,X)$ is a Kan-complex. But how do we proceed and see that every $\infty$-groupoid is indeed a Kan complex?

Answer 1

Sure. Suppose $X$ is an $\infty$-groupoid and let $B=Y=\Delta^0$ and $A=\emptyset.$ Then your theorem says that $K(\Delta^0,X)\to K(\emptyset, X)$ is a Kan fibration. But $K(\Delta^0,X)$ is isomorphic to $X,$ if $X$ is an $\infty$-groupoid, while $K(\emptyset,X)$ is isomorphic to $\Delta^0.$ Therefore the map $X\to\Delta^0$ is a Kan fibration and $X$ is a Kan complex as desired.

Better yet, your statement says that $K(\Delta^0,X)\cong X$ is a Kan complex directly!

This argument is totally contentless, of course; if you've proven your theorem then you've done the core work at some point elsewhere. The technical heart of this theorem is that $\infty$-groupoids allow fillings of "special outer horns", those outer horns whose extremal edge is invertible. This was Joyal's original approach to the theorem, which is important enough that it's often known as Joyal's Theorem. There are many expositions, such as around Theorem 2.1.8 in Land's recent Introduction to $\infty$-categories.