# abide law — how to say and generalization

1. Suppose that some algebraic operations $+$ and $\oplus$ satisfy the abide law, i.e. $(a_0+a_1)\oplus(b_0+b_1)=(a_0\oplus b_0)+(a_1\oplus b_1)$. How should I say this, “$+$ abides by $\oplus$” or what?
2. Is there a generalization of the abide law for any arity, i.e. $f(g(a_{00}, a_{01}, \ldots), g(a_{10}, a_{11}, \ldots), \ldots)$ $= g(f(a_{00}, a_{10}, \ldots), f(a_{01}, a_{11}, \ldots), \ldots)$?
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You should specify some context for the abide law, e.g. perhaps you are working with streams in functional programming languages, e.g. see Hinze: Concrete Stream Calculus. –  Bill Dubuque Jul 15 '11 at 21:27
@Bill Dubuque: Why? It's odd that we use different words in different contexts for the same concept. –  beroal Jul 18 '11 at 16:18

(1) The authors of these notes on polynomial relators would say that ‘$+$ abides $\oplus$’. However, the two operations appear symmetrically, a fact that is better reflected in the usages ‘$+$ and $\oplus$ abide’, ‘$+$ and $\oplus$ are abiding operations’ (as found here and here, for instance). However, the terminology seems to depend somewhat on the mathematical context: in category theoretic contexts I’ve seen this as ‘$+$ and $\oplus$ satisfy the interchange (or exchange) law’, e.g. here and in Saunders MacLane, Categories for the Working Mathematician. (This terminology is also mentioned in the notes on polynomial relators.)
(2) If $f$ and $g$ are $n$-ary operations for some $n$, the condition that$$f(g(a_{11},\dots,a_{1n}),\dots,g(a_{n1},\dots,a_{nn})) = g(f(a_{11},\dots,a_{n1}),\dots,f(a_{1n},\dots,a_{nn}))$$for all $n\times n$ matrices $[a_{ij}]$ for which the expressions are defined is certainly meaningful, but I have no idea whether it has found any use when $n>2$.
Actually, $f$ and $g$ may have different arity. @Brian M. Scott: –  beroal Jul 18 '11 at 16:24