First fix the following notations:

$AF:=$ The axiom of foundation

$ZFC^{-}:=ZFC\setminus \left\lbrace AF \right\rbrace $

$G:=$ The proper class of all sets

$V:=$ The proper class of Von neumann's cumulative hierarchy

$L:=$ The proper class of Godel's cumulative hierarchy

$G=V:~\forall x~\exists y~(ord(y) \wedge ``x\in V_{y}")$

$G=L:~\forall x~\exists y~(ord(y) \wedge ``x\in L_{y}")$

Almost all of $ZFC$ axioms have a same "nature" in some sense. They are "generating" new sets which form the world of mathematical objects $G$. In other words they are complicating our mathematical chess by increasing its playable and legitimated nuts. But $AF$ is an exception in $ZFC$. It is "simplifying" our mathematical world by removing some sets which innocently are accused to be "ill founded". Even $AF$ is regulating $G$ by $V$ and says $G=V$. So it is "miniaturizing" the "real" size of $G$ by the "very small" cumulative hierarchy $V$ as same as the assumption of constructibility axiom $G=L$. In fact "minimizing" the size of mathematical universe is ontological "nature" of all constructibilty kind of axioms like $G=W$ which $W$ is an arbitrary cumulative hierarchy. But in the opposite direction the large cardinal axioms says a different thing about $G$. We know that any large cardinal axiom stronger than "$0^{\sharp}$ exists" implies $G\neq L$. This illustrates the "nature" of large cardinal axioms. They implicitly say the universe of mathematical objects is too big and is "not" reachable by cumulative hierarchies. So it is obvious that any constructibility kind axiom such as $AF$, imposes a limitation on the height of large cardinal tree. One of these serious limitations is Kunen inconsistency theorem in $ZFC^{-}+AF$.

**Theorem (Kunen inconsistency)** There is no non trivial elementary embedding $j:\langle V,\in\rangle\longrightarrow \langle V,\in\rangle $ (or equivalently $j:\langle G,\in\rangle\longrightarrow \langle G,\in\rangle$)

The proof has two main steps as follows:

**Step (1):** By induction on Von neumann's "rank" in $V$ one can prove any non trivial elementary embedding $j:\langle V,\in\rangle\longrightarrow \langle V,\in\rangle$ has a critical point $\kappa$ on $Ord$.

**Step (2):** By iterating $j$ on this critical point one can find an ordinal $\alpha$ such that $j[\alpha]=\lbrace j(\beta)~|~\beta \in \alpha \rbrace \notin V (=G)$ which is a contradiction.

Now in the absence of $AF$ we must notice that the Kunen inconsistency theorem splits into two distinct statements and the orginal proof fails in both of them.

**Statement (1): (Strong version of Kunen inconsistency)** There is no non trivial elementary embedding $j:\langle G,\in\rangle\longrightarrow \langle G,\in\rangle$.

**Statement (2): (Weak version of Kunen inconsistency)** There is no non trivial elementary embedding $j:\langle V,\in\rangle\longrightarrow \langle V,\in\rangle$.

In statement (1), step (1) collapses because without $AF$ we have not a "rank notion" on $G$ and the induction makes no sense. So we can not find any critical point on $Ord$ for $j$ by "this method".

In statement (2), step (2) fails because without $AF$ we don't know $G=V$ and so $j[\alpha]\notin V$ is not a contradiction.

But it is clear that in $ZFC^{-}$ the original proof of Kunen inconsistency theorem works for both of the following propositions:

**Proposition (1):** There is no elementary embedding $j:\langle G,\in\rangle\longrightarrow \langle G,\in\rangle $ with a critical point on $Ord$.

**Proposition (2):** Every non trivial elementary embedding $j:\langle V,\in\rangle\longrightarrow \langle V,\in\rangle$ has a critical point on $Ord$.

Now the main questions are:

**Question (1):** Is the statement "There is a non trivial elementary embedding $j:\langle V,\in\rangle\longrightarrow \langle V,\in\rangle$" an acceptable large cardinal axiom in the absence of $AF$($G=V$)?
What about other statements by replacing $V$ with an arbitrary cumulative hierarchy $W$?(In this case don't limit the definition of a cumulative hierarchy by condition $W_{\alpha +1}\subseteq P(W_{\alpha})$)

Note that such statements are very similar to the statment "$0^{\sharp}$ exists" that is equivalent to existence of a non trivial elementary embedding $j:\langle L,\in\rangle\longrightarrow \langle L,\in\rangle$ and could be an "acceptable" large cardinal axiom in the "absence" of $G=L$. So if the answer of the question (1) be positive, we can go "beyond" weak version of Kunen inconsistency by removing $AF$ from $ZFC$ and so we can find a family of "Reinhardt shape" cardinals correspond to any cumulative hierarchy $W$ by a similar argument to proposition (2) dependent on "good behavior" of "rank notion" in $W$.

**Question (2):** Is $AF$ necessary to prove "strong" version of Kunen inconsistency theorem? In the other words is the statement "$Con(ZFC)\longrightarrow Con(ZFC^{-}+ \exists$ a non trivial elementary embedding $j:\langle G,\in\rangle\longrightarrow \langle G,\in\rangle)$" true?

It seems to go beyond Kunen inconsistency it is not necessary to remove $AC$ which possibly "harms" our powerful tools and changes the "natural" behavior of objects.It simply suffices that one omit $AF$'s limit on largeness of "Cantor's heaven" and his "set theoretic intuition". َAnyway whole of the set theory is not studying $L$, $V$ or any other cumulative hierarchy and there are many object "out" of these realms. For example without limitation of $G=L$ we can see more large cardinals that are "invisible" in small "scope" of $L$.In the same way without limitation of $AF$ we can probably discover more stars in the mathematical universe out of scope of $V$. Furthermore we can produce more interesting models and universes and so we can play an extremely exciting mathematical chess beyond inconsistency, beyond imagination!