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I am proving that $\mathtt{NAND}$ is functionally complete in Boolean algebra using Robert Veroff’s 2-basis for the Sheffer stroke ($\mathtt{NAND}$) from his paper (A Shortest 2-Basis for Boolean Algebra in Terms of the Sheffer Stroke). I need clarification on how two derived identities from the paper’s proof contribute to constructing the standard $\mathtt{NAND}$ expressions for basic Boolean operations to establish functional completeness.

Veroff’s 2-basis is:

  1. Commutativity: $x \mid y = y \mid x$
  2. Main axiom: $\left( x \mid y \right) \mid \left( x \mid \left( y \mid z \right) \right) = x$

The two derived identities are:

$\blacklozenge \quad \left( x \mid y \right) \mid \left( y \mid y \right) = y \quad$ (step 22, page 4)

$\lozenge \quad x \mid \left( y \mid \left( y \mid y \right) \right) = x \mid x \quad$ (step 64, page 5)

To prove $\mathtt{NAND}$ is functionally complete, I need to express all Boolean operations (e.g., $\mathtt{NOT}$, $\mathtt{AND}$, $\mathtt{OR}$) using only $\mathtt{NAND}$. The standard expressions are:

  • $\mathtt{NOT}\left( x \right) = x \mid x$, which follows directly from the $\mathtt{NAND}$ definition: $x \mid x = \neg \left( x \wedge x \right) = \neg x$.
  • $\mathtt{AND}\left( x, y \right) = \left( x \mid y \right) \mid \left( x \mid y \right)$, since $\left( x \mid y \right) \mid \left( x \mid y \right) = \neg \left( x \wedge y \right) \mid \neg \left( x \wedge y \right) = x \wedge y$.
  • $\mathtt{OR}\left( x, y \right) = \left( x \mid x \right) \mid \left( y \mid y \right)$, since $\left( x \mid x \right) \mid \left( y \mid y \right) = \neg x \mid \neg y = \neg \left( \neg x \wedge \neg y \right) = x \vee y$.

Question: How do the derived identities $\left( x \mid y \right) \mid \left( y \mid y \right) = y$ (step 22) and $x \mid \left( y \mid \left( y \mid y \right) \right) = x \mid x$ (step 64) support the derivation of the standard $\mathtt{NAND}$ expressions for $\mathtt{AND}\left( x, y \right) = \left( x \mid y \right) \mid \left( x \mid y \right)$ and $\mathtt{OR}\left( x, y \right) = \left( x \mid x \right) \mid \left( y \mid y \right)$ using Veroff’s 2-basis? Can these identities, combined with the 2-basis, be used in paramodulation or other equational reasoning steps to derive these expressions? If they do not directly contribute to these derivations, how do they help prove that the 2-basis ensures $\mathtt{NAND}$’s functional completeness?

The paper shows the 2-basis is equivalent to Sheffer’s 3-basis, implying all Boolean operations are expressible. However, I need to understand the specific role of steps 22 and 64 in enabling the construction of $\mathtt{AND}$ and $\mathtt{OR}$, possibly in conjunction with other derived identities (e.g., step 17: $\left( x \mid x \right) \mid \left( x \mid y \right) = x$, or step 26: $\left( x \mid x \right) \mid y = y \mid \left( y \mid x \right)$).

Dang Dang
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1 Answers1

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Sheffer, in his original article, proposed three equational axioms describing his operation $|$.

These are given in §1.III as 3,4,5, phrased in terms of both $|$ and the defined notation $x'$ abbreviating $x|x$.

In the rest of the article, using these 3 laws, Sheffer proves that the defined operations $x'$, $(x|y)'$ and $(x'|y')$ satisfy the same laws as the usual Boolean algebra operations in the axiom system of Huntington.

So all Veroff needs to show is that his two axioms imply the three axioms of Sheffer: then the completeness of his axiom system follows from Sheffer's result. Veroff's 22, 64 and 86 allow him to do this: the latter two correspond directly to Sheffer's 4 and 5, whereas Veroff's 22 gives Sheffer's 3 by taking $x=y$.

Z. A. K.
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