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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.


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. $y^\bot \leq x^\bot$
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No.

Hexagon

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?

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    $\begingroup$ Good answer. I'll let you know if I have any ideas. $\endgroup$ – goblin Dec 11 '13 at 3:05
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    $\begingroup$ Yes, they imply orthomodularity and even distributivity, so $L$ must be a Boolean algebra. Indeed, by the first shuffle law, $L$ must be a Heyting algebra/Brouwerian lattice, which is necessarily distributive. $\endgroup$ – Tristan Bice Feb 10 '16 at 23:13

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