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Consider the structure $(\{0,1\}, *)$, where $*$ represents the conditional operator in logic, defined by $1*1=0*0=0*1=1$ and $1*0=0$. Is there a finite basis for the equational identities of that algebraic structure, and if so, can someone exhibit a finite list of identities?

user107952
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1 Answers1

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From R. Lyndon, Identities in two-valued calculi, Trans Amer Math Soc 71 (1951), 457-465, we know that every 2-element algebraic structure in a finite language has a finite basis of equational identities.

From J. C. Abbott, Implicational algebras, Bulletin mathématique de la Société des Sciences Mathématiques de la République Socialiste de Roumanie Nouvelle Série, Vol. 11 (59), (1967), 3-23, we know an explicit basis of equational identities for the 2-element implication algebra $\langle \{0,1\}; \to \rangle$:

  • $((x\to y)\to x) = x$,
  • $((x\to y)\to y) = ((y\to x)\to x)$,
  • $(x\to (y\to z))=(y\to (x\to z))$.
  • Keith Kearnes
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