Does every orthocomplemented lattice satisfy the shuffle laws?

Question

Let $L=(X,\wedge,\vee,^\bot)$ denote a non-empty lattice having a unary operation $x \mapsto x^\bot$ that satisfies both "shuffle" laws: $$x \wedge y \leq z \iff x \leq y^\bot \vee z.$$

$$x \leq y \vee z \iff x \wedge y^\bot \leq z.$$

Then $L$ is an orthocomplemented lattice. (Proof below).

Question. Does every orthocomplemented lattice satisfy the shuffle laws? It suffices to prove just one of them, but I've been unable to do so.

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Proof of claim.

Complements. Let $x$ be fixed but arbitrary. This is allowed, since $L$ was assumed non-empty. We will show that $x^\bot \wedge x$ is a least element. TFAE.

1. $x^\bot \wedge x \leq a$
2. $x^\bot \leq x^\bot \vee a$
3. TRUE

Thus $x^\bot \wedge x$ is a least element. A similar argument shows that $x^\bot \vee x$ is a maximum element.

Involutiveness. We will show that $x \leq x^{\bot\bot}.$ TFAE.

1. $x \leq a$
2. $x \wedge x \leq a$
3. $x \leq x^\bot \vee a$
4. $x \wedge x^{\bot\bot} \leq a$

So the elements above $x$ and $x \wedge x^{\bot\bot}$ are identical; by antisymmetry, $x = x \wedge x^{\bot\bot}$. It follows that $x \leq x^{\bot\bot}.$

A similar argument shows the other direction; therefore $x = x^{\bot\bot}.$

Order-reversingness. The following are equivalent.

1. $x \leq y$
2. $x \wedge 1 \leq y$
3. $1 \leq x^\bot \vee y$
4. $1 \wedge y^\bot \leq x^\bot$
5. $y^\bot \leq x^\bot$

Answer 1

No.

Put $x=a^\perp$, $y=b^\perp$, $z=b$. The first shuffle law fails.

But this makes the whole thing more interesting. Maybe the shuffle laws imply orthomodularity?