Robbins Algebra

A Robbins algebra is a nonempty set A satisfying the following four axioms, expressed in clause notation, in which the function o can be interpreted as plus and the function n as negation.

(R1) EQ(o(x,y),o(y,x)).	(R3) EQ(n(o(n(o(x,y)),n(o(x,n(y))))),x).
(R2) EQ(o(o(x,y),z),o(x,o(y,z))).	EQ(x,x).

A Boolean algebra is a nonempty set S with two operations, plus and times, and a 1 and a 0. Each operation is commutative, and each distributes over the other. The 1 is a multiplicative identity, and the 0 is an additive identity. In addition, for every x, the negation of x exists with x plus its negation equal to 1 and x times its negation equal to 0. An alternative axiomatization of Boolean algebra consists of (R1), (R2), and Huntington's axiom (H3) [Huntington33].

(H3) EQ(o(n(o(n(x),y)),n(o(n(x),n(y)))),x).

Whether Robbins implies Boolean is still an open question. What is known is that the addition to the three Robbins axioms of any one of a number of properties of a Boolean algebra suffices to yield Boolean.

Problem RA1, simple. Prove that, if the following axiom is adjoined to the axioms for a Robbins algebra, the resulting algebra is Boolean. We recommend trying to prove Huntington's axiom (H3).

EQ(o(x,0),x).

Problem RA2, moderate. Prove that t addition of the following equality to Robbins yields Boolean.

EQ(o(x,x),x).

Problem RA3, hard. Prove that, where c is a constant, the addition of the following equality to Robbins yields Boolean.

EQ(o(c,c),c).

Problem RA4, never proved in a single run unaided. Prove that, where c and d are constants, the addition of the following equality to Robbins yields Boolean.

EQ(o(c,d),d).

Problem RA5, never proved in a single run unaided. Prove that, where c and d are constants, the addition of the following equality to Robbins yields Boolean.

EQ(n(o(c,d)),n(d)).