I am convinced in the existence of large cardinals by the presence of many near inconsistencies in the large cardinal hierarchy without any actual inconsistency (and for other reasons as well that others have elaborated on elsewhere). I have written computer programs that search for inconsistencies in the large cardinal hierarchy and which easily produce many near inconsistencies. I see no explanation for these near inconsistencies except for the existence of very large cardinals. If large cardinals exist, then it is possible that they behave in exotic ways that look to us as if the theory of large cardinals is always on the verge of an inconsistency. But if large cardinals do not exist, then we do not have an explanation for so many near inconsistencies.
Algebras of rank-into-rank embeddings:
Let $\lambda$ be a cardinal, and let $\mathcal{E}_\lambda$ be the set of all elementary embeddings $j:V_\lambda\rightarrow V_\lambda$. Then recall that $\mathcal{E}_\lambda$ may be endowed with an algebraic operation $*$ defined by
$j*k=\bigcup_{\alpha<\lambda}j(k|_{V_\alpha})$. If $\gamma<\lambda$ and $\gamma$ is a limit ordinal, then define a congruence $\equiv^\gamma$ on $(\mathcal{E}_\lambda,*,\circ)$ by setting $j\equiv^\gamma k$ iff $j(x)\cap V_\gamma=k(x)\cap V_\gamma$ whenever $x\in V_\gamma$. The algebra $\mathcal{E}_\lambda$ satisfies the self-distributivity identity $j*(k*l)=(j*k)*(j*l)$
We say that an algebra $(X,*,1)$ is a reduced Laver-like algebra if it satisfies the self-distributivity identity $x*(y*z)=(x*y)*(x*z)$ along with $x*1=1,1*x=x$ and where if $x_n\in X$ for each $n\geq 0$, then $x_0*\dots*x_N=1$ for some $N$ (we abide by the convention that the implied parentheses are grouped on the left so that $a*b*c*d=((a*b)*c)*d$). The algebra $\mathcal{E}_\lambda/\equiv^\gamma$ is always a reduced Laver-like algebra, but there exists an 11 element Laver-like algebra that does not embed in any $\mathcal{E}_\lambda/\equiv^\gamma$.
If $(X,*,1)$ is a Laver-like algebra, then define an operation $*_n$ for all $n\geq 0$ by setting $x*_0y=y,x*_{n+1}y=x*(x*_ny)$ for all $n$ and set $x*_\infty y=\lim_{n\rightarrow\infty}x*_ny$. Then we say that $\text{crit}(x)\leq\text{crit}(y)$ precisely when $x*_\infty y=1$. The set $\text{crit}[X]=\{\text{crit}(x):x\in X\}$ is always a linearly ordered set.
Suppose that $(X,*,1)$ is a reduced Laver-like algebra. Then define an equivalence relation $\simeq_{\text{cmx}}$ on $X$ by setting $x\simeq_{\text{cmx}}y$ if and only if $x*a_1*\dots*a_r=1\Leftrightarrow y*a_1*\dots*a_r=1$ for all $a_1,\dots,a_r\in X$ and $r\geq 0$. Then $\simeq_{\text{cmx}}$ is a congruence. We say that two Laver-like algebras with generating sets $(X,*,1,(x_a)_{a\in A})$ and $(Y,*,1,(y_a)_{a\in A})$ are critically equivalent if
$(X,*,1,(x_a)_{a\in A})/\simeq_{\text{cmx}}$ is isomorphic to $(Y,*,1,(y_a)_{a\in A})/\simeq_{\text{cmx}}.$
For many purposes, we only care about reduced Laver-like algebras up to critical equivalence.
I have developed an algorithm that takes a finite Laver-like algebra $(X,*,1)$ with finite generating set $(x_a)_{a\in A}$ as an input and returns all algebras with generating sets $(Y,*,1,(y_a)_{a\in A})$ up-to-critical equivalence such that $|\text{crit}[Y]|=|\text{crit}[X]|+1$ and where there is a surjective homomorphism $\phi:Y\rightarrow X$ with $\phi(y_a)=x_a$ for all $a.$ In other words, this algorithm takes a finite Laver-like algebra as an input and returns all Laver-like algebras with one more critical point up to critical equivalence but the same generating set. By repeatedly applying this algorithm, one can exhaustively search for Laver-like algebras that satisfy various conditions.
Suppose that $P$ is a property of Laver-like algebras and that if $(X,*,1),(Y,*,1)$ are critically equivalent Laver-like algebras, then $(X,*,1)$ satisfies $P$ if and only if $(Y,*,1)$ satisfies $P$. Suppose furthermore that large cardinals imply the existence of a subalgebra of some $\mathcal{E}_\lambda/\equiv^\gamma$ that satisfies $P$. Then if we exhaustively search for algebras up-to-critical-equivalence that satisfy Property $P$ but we turn up empty handed, then we have obtained an inconsistency in the large cardinal hierarchy. I have never observed such an inconsistency (except for the time where there was a bug in the code). But if an exhaustive search turns up just one algebra that satisfies Property $P$ up-to-critical equivalence, then we have obtained a near inconsistency.
Consistency tests:
Here are some ways to produce new Laver-like algebras from old ones. One can then perform an exhaustive search to make sure that these Laver-like algebras really exist or whether we have an inconsistency.
Test 1: Roots of algebras. Suppose that $t$ is a term in the language with function symbols $*,\circ$. Let $j_1,\dots,j_r:V_{\lambda+1}\rightarrow V_{\lambda+1}$ be elementary embeddings. Then by elementarity $\exists x_1,\dots,x_r\in\mathcal{E}_\lambda,t(x_1,\dots,x_r)*x_s=j_s|_{V_\lambda}$ if and only if $\exists x_1,\dots,x_r\in\mathcal{E}_\lambda,t(x_1,\dots,x_r)*x_s=t(j_1,\dots,j_r)*j_s|_{V_\lambda}$ which is true. Therefore $\exists x_1,\dots,x_r\in\mathcal{E}_\lambda,t(x_1,\dots,x_r)*x_s=j_s|_{V_\lambda}$. We can use these new elementary embeddings to produce Laver-like algebras, but an exhaustive backtracking search of these Laver-like algebras may verify that these Laver-like algebras actually exist.
For the following tests, let $\text{crit}_n(j_1,\dots,j_r)$ be the $n$-th element of the set $\{\text{crit}(k):k\in\langle j_1,\dots,j_r\rangle\}$ (which has order type $\omega$).
Test 2: Adding an ordinal. Suppose that $g$ is a well-ordering of $V_\lambda$ where if $\text{Rank}(x)<\text{Rank}(y)$, then $g(x)<g(y)$. Let $\mathcal{E}_\lambda[g]$ be the set of all elementary embeddings $j$ in $\mathcal{E}_\lambda$ where
$j(g(x))=g(j(x))$ for all $x$. The non-emptyness of $\mathcal{E}_\lambda[g]\setminus\{1_{V_\lambda}\}$ follows whenever $g$ is a $V_\lambda$-definable well-ordering which exists whenever $V_\lambda\models V=HOD$, but there is a simpler argument for the non-emptyness that holds in all models.
Suppose now that $j_1,\dots,j_r\in\mathcal{E}_\lambda[g]\setminus\{1_{V_\lambda}\}$. Let $\gamma=\text{crit}_n(j_1,\dots,j_r)$. Then since $\langle j_1,\dots,j_r\rangle/\equiv^\gamma$ is finite, there are sets $A_1,\dots,A_s\in V_\gamma$ where if $j,k\in\langle j_1,\dots,j_r\rangle$, then $j\not\equiv^\gamma k$, then $j(A_s)\cap V_\gamma\neq k(A_s)\cap V_\gamma$ for some $s$. Therefore, if we set $A=\{A_1,\dots,A_s\}$, then whenever $j,k\in\langle j_1,\dots,j_r\rangle$, if $j\not\equiv^\gamma k$, then $j(A)\neq k(A)$. Therefore, if we set $\alpha=g(A)\cdot\omega$, then $\alpha<\gamma$ but if
$j\not\equiv^\gamma k$, then $j(\alpha)\neq k(\alpha)$. The ordinal $\alpha$ may be treated as an element in a Laver-like algebra that extends $\mathcal{E}_\lambda/\equiv^\delta$, but we can exhaustively search for Laver-like algebras that have isomorphic copies of the ordinal $\alpha$.
Test 3: Extending a generator.
With rank-into-rank embeddings, one can always extend the algebra
$\langle j_1,\dots,j_r\rangle/\equiv^{\text{crit}_n(j_1,\dots,j_r)}$ to the larger algebra $\langle j_1,\dots,j_r\rangle/\equiv^{\text{crit}_{n+1}(j_1,\dots,j_r)}$, but we can do even better by taking more control over the extension.
Proposition: Suppose that $j_1,\dots,j_r\in\mathcal{E}_\lambda\setminus\{1_{V_\lambda}\}$ and $n$ is a natural number. Then there are $k_1,\dots,k_r\in\langle j_1,\dots,j_r\rangle$ where
$k_s\equiv^{\text{crit}_n(j_1,\dots,j_r)}j_s$ for all $s$ but where if
$\text{crit}(k_{a_1}*\dots*k_{a_s})=\text{crit}_n(k_1,\dots,k_r)$ and
$\text{crit}(k_{a_1}*\dots*k_{a_t})<\text{crit}_n(k_1,\dots,k_r)$ for all $1\leq t<s$, then $a_1=1$.
Some data
We say that a Laver-like algebra $(X,*,1)$ with generating set $(x_a)_{a\in A}$ is superreduced if $x*x=x*y=y*x=y*y=1$ implies that $x=y$, and if $(X,*,1,(x_a)_{a\in A})$ is superreduced, then we say that $(X,*,1,(x_a)_{a\in A})$ is a superreduced multigenic Laver table if whenever $(Z,*,1,(z_a)_{a\in A})$ is critically equivalent to $(X,*,1,(x_a)_{a\in A})$, there is a surjective homomorphism $\phi:X\rightarrow Z$ with $\phi(x_a)=z_a$ for $a\in A$. Let $\mathcal{S}$ be a class of superreduced multigenic Laver tables such that if $(X,*,1,(x_a)_{a\in A})$ is a superreduced multigenic Laver table, then there is a unique $(Y,*,1,(Y_a)_{a\in A})\in S$ isomorphic to $(X,*,1,(x_a)_{a\in A})$ (we do not need the axiom of choice to construct $\mathcal{S}$ since there is a standard construction of $\mathcal{S}$). Then let's call the elements of $\mathcal{S}$ standard multigenic Laver tables.
Suppose that $(X,*,1)$ is a standard superreduced Laver-like algebra generated by $(x_a)_{a\in A}$. For each string $g$ over the alphabet $A$, let $\text{Ext}(g,(X,*,1,(x_a)_{a\in A}))$ be the set of all standard superreduced multigenic Laver tables where we set $\text{Ext}(\epsilon,(X,*,1,(x_a)_{a\in A}))=\{(X,*,1,(x_a)_{a\in A})\}$ and where if $g$ is a string in $A^*$ and $a_0\in A$, then $\text{Ext}(ga_0,(X,*,1,(x_a)_{a\in A}))$ is the set of all standard superreduced multigenic Laver tables $(Z,*,1,(z_a)_{a\in A})$ where there is some $(Y,*,1,(y_a)_{a\in A})\in \text{Ext}(ga_0,(X,*,1,(x_a)_{a\in A}))$ and where if $\text{crit}(z_{a_1}*\dots*z_{a_s})=\max(\text{crit}[Z]\setminus\{\text{crit}(1)\})$ and $\text{crit}(z_{a_1}*\dots*z_{a_t})<\text{crit}(z_{a_1}*\dots*z_{a_s})$ whenever $1\leq t<s$, then $a_1=a_0$, and where $|\text{crit}[Z]|=|\text{crit}[Y]|+1$ and there is a surjective homomorphism $\phi:Z\rightarrow Y$ with $\phi(z_a)=y_a$ for all $a\in A$.
Let $A=\{a,b,c\}$. Let
$(X,*,1,(x_a)_{a\in A})$ be the standard superreduced multigenic Laver table with $|X/\simeq_{\text{cmx}}|=3$ and $\text{crit}(x_a)<\text{crit}(x_b)<\text{crit}(x_c)$. Let $\mathcal{A}$ be the collection of strings $g$ over the alphabet $A$ where if $h$ is a prefix of $g$, then $|\text{Ext}(h,(X,*,1,(x_a)_{a\in A}))|=1$ and where if $Y\in\text{Ext}(g,(X,*,1,(x_a)_{a\in A}))$, then $|Y|\leq 54$ and where $\text{crit}(x*x*y)\leq\text{crit}(x*y)$ for all $x,y\in Y$. Then there are 15215 elements in $\mathcal{A}$. I have computed $\mathcal{A}$ using one CPU core in 220 seconds.
Orthogonality
In the above example, many of the near inconsistencies that I have found are similar to one another. To see why this is the case, we will have to look at some non-commutative polynomials. If $(X,*,1)$ is a Laver-like algebra with generating set $(x_a)_{a\in A}$, and $\alpha$ is a critical point, then define a non-commutative polynomial $p_\alpha((z_a)_{a\in A})=\{z_{a_1}\dots z_{a_r}:\text{crit}(x_{a_1}*\dots x_{a_r})=\alpha,\forall s<r,\text{crit}(x_{a_1}*\dots x_{a_r})<\alpha\}.$ Let $M(X,*,1,(x_a)_{a\in A})=p_{\beta}((z_a)_{a\in a})$ where $\beta=\max(\text{crit}[X]\setminus\{\text{crit}(1)\})$.
Then $|M[\mathcal{A}]|=65$. This indicates that the 15215 near inconsistencies are all very similar to one another. Fortunately, by combining several inconsistency tests together, one can achieve a far more diverse collection of near inconsistencies.
Conclusion
I have not done much research on these kinds of near inconsistencies (and neither has anyone else), so there is a meager possibility that there is an explanation for these near inconsistencies or a theorem in ZFC that explains why we are seeing these near inconsistencies. There is also the slight possibility that these near inconsistencies arise in a universe without very large cardinals because a similar structure that also produces these near inconsistencies exists (this could be an inner model or an algebraic structure).
