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1868 Nov. 3.
[????] [let?]
lx
denote that whatever is l to every x.
Then 1(x+,y) = l(x), l(y) unless either x or y = 0
But if we let
lx
denote whatever is l to some x
Then l(x+,y) = l(x) +, l(y)
Turn over.