MATH.2:5.3 - Identifying words changes the question they answer

Take the one-generator paths a^n from MATH.1, with concatenation a^m;a^n=a^(m+n). Impose a^2~a^0 and choose the smallest equivalence relation compatible with concatenation that contains this equation.

Compatibility propagates this equation under concatenation. Adding one a gives a^3~a, and repeated deletion of a pair reduces every even-length word to the empty path and every odd-length word to a. These two groups remain distinct: parity itself is a compatible relation satisfying the imposed equation, as :5.1 shows, so the smallest such relation cannot identify opposite parities. This yields two classes and a composition table identical to the addition table above.

The quotient can describe the parity of repeated toggling. It discards the number of toggles. If each use takes time, elapsed cost cannot be recovered from those two classes alone. Retain the length or accumulated cost for a question that consumes it.