Survey 1979: Equational Logic - Department of Mathematics ...

16 EQUATIONAL LOGIC
(Martin [286, page 56]). (Here co= { 0,1,2,...} and xY denotes ordinary
exponentiation with 00 = 1.) The proof is surprisingly long. Cf. 9.20 below.
8.11. The algebra (co,xy,x y) is generic for the laws
(Martin [286, page 78] ).
xy = yx
(xy)z = x(yz)
(xy)Z = xZy z
(xY)Z = xYZ
8.12. The algebra (FZ,+) is generic for the laws
(x+y) + z = x + (y+z)
x+y+x+y=y+x+x+y
x+y+z+x+y=y+x+z+x+y
x+y+x+z+y=y+x+x+z+y
x+y+z+x+w+y=y+x+z+x+w+y
(J. Karnofsky [unpublished] - see [286, page 31]). Here + denotes addition on the
class FZ of all ordinals - the algebra (FZ,+) could be replaced by a countable one.
8.13. The algebras (co;An)n>3 and (co;0n)n>4 are each generic for the variety of
all algebras of type (2,2,2,..) (i.e., for Z0 = 0). (Martin [286, page 131, page 134] .)
Here A n (n > 3)are the Ackermann operations beyond exponentiation, and the O n are
some related operations invented by Doner and Tarski [110], who conjectured a
somewhat stronger statement.
algebras?
PROBLEM 1. Do these three equations form an axiom base for Boolean
xvy=yvx
xv(yvz)=(xvy)vz
((x v y)'v (x v y')')'= x.
(See [354] for a history of this problem.) All finite models of these equations are
Boolean algebras.
PROBLEM 2. Do the following eleven equations form an axiom base for
(co,1 ,x+y,xy,xY)?
1. x+y=y+x