Even if solipsism is true, it is unknowable

The proposition expressed by the title is what I've found recently. Anyone who has or knows similar ideas please let me know. Other comments are also welcome.

A paraphrase of solipsism: no one is possible to know what anyone else knows. More precisely, for any $x$ and $y$, if $x$ is not $y$ then it is impossible that $x$ knows what $y$ knows. In formula: \[\forall x\forall y(x\neq y\rightarrow\forall p\neg\Diamond(K_{x}K_{y}p\lor K_{x}\neg K_{y}p))\]

Let's denote the formula by $S$. Assuming

(a) $\vdash\Box\varphi\rightarrow\varphi$, we have \[\forall x\forall y(x\neq y\rightarrow\forall p(\neg K_{x}K_{y}p\land\neg K_{x}\neg K_{y}p))\]

Let's denote it by $S'$. Thus $\vdash S\rightarrow S'$. Suppose $a\neq b$. From $S'$ it follows that

(1) $\forall p(\neg K_{a}K_{b}p\land\neg K_{a}\neg K_{b}p)$, and

(2) $\forall p(\neg K_{b}K_{a}p\land\neg K_{b}\neg K_{a}p)$.

From (1) we have

(3) $\neg K_{a}K_{b}p\land\neg K_{a}\neg K_{b}p$, which implies that

(4) $\neg K_{a}K_{b}p$

From (2) by substituting $K_{b}p$ for $p$ we have

(5) $\neg K_{b}K_{a}K_{b}p\land\neg K_{b}\neg K_{a}K_{b}p$, which implies that

(6) $\neg K_{b}\neg K_{a}K_{b}p$ Let's denote $\neg K_{a}K_{b}p$ by $P$. Then we have

(7) $\vdash S\rightarrow P$, and

(8) $\vdash S\rightarrow\neg K_{b}P$ Assuming the rule

(b) $\vdash \varphi\rightarrow\psi\Longrightarrow\ \vdash K\varphi\rightarrow K\psi$,

from (7) we have

(9) $\vdash K_{b}S\rightarrow K_{b}P$. Assuming

(c) $\vdash K\varphi\rightarrow\varphi$, it follows from (8) that

(10) $\vdash K_{b}S\rightarrow\neg K_{b}P$ Now (9) and (10) together imply that

(11) $\vdash\neg K_{b}S$. Assuming the rule

(d) $\vdash\varphi\Longrightarrow\ \vdash\Box\varphi$, we have

(12) $\vdash\neg\Diamond K_{b}S$ Note that $b$ can be arbitrary. Therefore,

(13) $\vdash\forall x\neg\Diamond K_{x}S$, which means that $S$ is unknowable. In arriving at the conclusion, we only assumed (a)--(d), which are all plausible for modal logic and epistemic logic.