Let me expound on a somewhat plausible scenario of an inconsistency in the large cardinal hierarchy that may take a very long time to appear.

A rank-into-rank embedding is an elementary embedding $j:V_{\lambda}\rightarrow V_{\lambda}$.

**Lemma** If $j$ is a rank-into-rank embedding, then $(j*j)(\alpha)\leq j(\alpha)$ for all ordinals $\alpha$.

**Proof** Suppose that $\alpha<\lambda$. Let $\beta$ be the least ordinal such that $j(\beta)>\alpha$. Then
$$V_{\lambda}\models\forall x<\beta,j(x)\leq\alpha.$$
Therefore, by elementarity,
$$V_{\lambda}\models\forall x<j(\beta),(j*j)(x)\leq j(\alpha).$$
In particular, since $\alpha<j(\beta)$, we conclude that $(j*j)(\alpha)\leq j(\alpha)$. $\mathbf{QED}.$

The above lemma has the following corollary as a purely combinatorial consequence.

If $1\leq x\leq 2^{n}$, then let $o_{n}(x)$ be the least natural number such that if $A_{n}$ is the classical Laver table of cardinality $2^{n}$, then $A_{n}\models x*2^{o_{n}(x)}=2^{n}$. In other words, $o_{n}(x)=\text{log}_{2}(p)$ where $p$ is the period of $x$ in the classical Laver table $A_{n}$.

**Corollary** Assume there exists a rank-into-rank cardinal. Then for all natural numbers $n$, one has $o_{n}(1)\leq o_{n}(2)$.

The functions $n\mapsto o_{n}(1)$ and $n\mapsto o_{n}(2)$ are strictly increasing. However, Randall Dougherty has shown that these functions increase very very slowly. A straightforwards computer calculation shows that $o_{n}(1)\leq o_{n}(2)$ whenever $o_{n}(1)\leq 4$.

Under large cardinal hypotheses, we have $o_{n}(1)\rightarrow\infty$. However, the statement $o_{n}(1)\rightarrow\infty$ does not seem to imply that $o_{n}(1)\leq o_{n}(2)$ for all $n$. In the following theorem, $f_{n}^{\text{Ack}}$ is a version of the Ackermann function.

**Theorem (Dougherty)** The least natural number $n$ such that $o_{n}(1)\geq 5$ is at least $f_{9}^{\text{Ack}}(f_{8}^{\text{Ack}}(f_{8}^{\text{Ack}}(254)))$.

**Corollary** If there is a natural number $n$ with $o_{n}(1)>o_{n}(2)$, then the least natural number $n$ such that $o_{n}(1)>o_{n}(2)$ is greater than $f_{9}^{\text{Ack}}(f_{8}^{\text{Ack}}(f_{8}^{\text{Ack}}(254)))$.

Therefore if one tries to prove that rank-into-rank cardinals are inconsistent by exhibiting an $n$ such that $o_{n}(1)>o_{n}(2)$, then one would need to take more than $f_{9}^{\text{Ack}}(f_{8}^{\text{Ack}}(f_{8}^{\text{Ack}}(254)))$ steps.

Of course, there may be short-cuts in showing that $o_{n}(1)>o_{n}(2)$ for some $n$ which take much less time than actually calculating the least $n$ such that $o_{n}(1)=5$. Or there could be an entirely different sort of contradiction with the assertion that there exists a rank-into-rank cardinal.

For the record, here are the first few values of $o_{n}(1)$ and $o_{n}(2)$.

$o_{1}(1)=0,o_{1}(2)=1;o_{2}(1)=1,o_{2}(2)=1;o_{3}(1)=2,o_{3}(2)=2;o_{4}(1)=2,o_{4}(2)=2;o_{5}(1)=3,o_{5}(2)=3;o_{6}(1)=3,o_{6}(2)=3;o_{7}(1)=3,o_{7}(2)=4;o_{8}(1)=3,o_{8}(2)=4;o_{9}(1)=4,o_{9}(2)=4$.

If such an inconsistency were to pop up only very far away, then such an inconsistency will not have any effect on the mathematics that mathematicians here on Earth will do because nobody will live long enough to observe such a contradiction.

This being said, set theorists generally do not think that there is any contradiction anywhere in the large cardinal hierarchy up until say $I0$ no matter how distant the contradiction is located.