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For an example, consider the statement $A \land (B \lor C)$.

The proposition $A$ distributes over $(B \lor C)$, so the statement is logically equivalent to $(A \land B) \lor (A \land C)$

Wikipedia claims that distributivity isn't a characteristic of quantum logic

Quantum logic has some properties that clearly distinguish it from classical logic, most notably, the failure of the distributive law of propositional logic.

If that's true, then according to the rules of quantum logic $A \land (B \lor C) \nRightarrow (A \land B) \lor (A \land C)$

Does that example, by itself, express a difference between the syntax of classical logic and quantum logic or does it express a difference between the semantics of those logics?

Hal
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1 Answers1

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You can see :

  • Kurt Engesser & Dov Gabbay & Daniel Lehmann (editors), Handbook of Quantum Logic and Quantum Structures : Quantum Logic (2008), Ch.1 The Birkhoff–von Neumann Concept of Quantum Logic by Miklos Rédei, page 1-on :

Let $\mathcal K = \{ P, \&, \sqcup, \sim \}$ be a zeroth order formal language with the set $P$ of sentence variables $p, q, \ldots,$ two place connectives $\&$ (and), $\sqcup$ (or), negation sign $\sim$, parentheses (,), and let $\mathcal F$ be the set of well formed formulas in $\mathcal K$ defined in the standard way by induction from $P$ : $\mathcal F$ is the smallest set for which the following two conditions hold:

$P ⊂ \mathcal F$ (1)

<p>if $\phi, \psi \in \mathcal F$, then $(\phi \&amp; ψ), (\phi \sqcup ψ), (\sim \phi) \in \mathcal F$ (2)</p>

Let $(\mathcal L,∨,∧,⊥)$ be an orthomodular lattice. Orthomodularity of $\mathcal L$ means that the following condition holds:

(3) orthomodularity: If $A ≤ B$ and $A^⊥ ≤ C$, then $A∨(B∧C) = (A∨B)∧(A∨C)$.

Orthomodularity is a weakening of the modularity law:

(4) modularity: If $A ≤ B$, then $A ∨ (B ∧ C) = (A ∨ B) ∧ (A ∨ C)$

which itself is a weakening of the distributivity law:

(5) distributivity: $A ∨ (B ∧ C) = (A ∨ B) ∧ (A ∨ C)$ for all $A,B,C$.

The set $\mathcal G_Q ⊆ \mathcal L$ in an orthomodular lattice is called a generalized filter if

if $A \in \mathcal G_Q$ and $A^⊥ ∨ (A ∧ B) \in \mathcal G_Q$ then $B \in \mathcal G_Q$ (7).

Given a pair $(\mathcal L, \mathcal G_Q)$ the map $i : \mathcal F \to \mathcal L$ is called an $(\mathcal L, \mathcal G_Q)$-interpretation if

$i(\phi \& ψ) = i(\phi) ∧ i(ψ)$ (8)

<p>$i(\phi \sqcup ψ) = i(\phi) ∨ i(ψ)$ (9)</p>

<p>$i(\sim \phi) = i(\phi)^⊥$ (10).</p>

Each interpretation $i$ determines a $(\mathcal L, \mathcal G_Q)$-valuation $v_i$ by :

(11) $v_i(\phi) = 1 \ \text {(true) if} \ \ i(\phi) \in \mathcal G_Q$,

<p>$ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0 \ \text {(false) if} \ \ i(\sim \phi) \in \mathcal G_Q$,</p>

<p>undetermined otherwise.</p>

If $V (\mathcal L)$ denotes the set of all $(\mathcal L, \mathcal G_Q)$-valuations and $V$ is the class of valuations determined by the class of orthomodular lattices, then $\phi \in \mathcal F$ is called valid if $v(\phi) = 1$ for every $v \in V$, and a class of formulas $\Gamma$ is defined to entail $\phi$ if $v(ψ) = 1$ for all $ψ \in \Gamma$ implies $v(\phi) = 1$.

One can define the quantum analog $\to_Q$ of the classical conditional by

(12) $\phi \to_Q ψ = \sim \phi \sqcup (\phi \& ψ)$

and one can formulate a deduction system in $\mathcal K$ using $\to_Q$ such that one can prove soundness and completeness theorems for the resulting quantum logical system.