* my formula again.
@ 1997-07-01 18:11 categories
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Date: Tue, 1 Jul 1997 11:05:19 +0100 (BST)
From: Paul Taylor <pt@dcs.qmw.ac.uk>
Thanks to Pino Rosolini for pointing out a typo in my formula:
for any function f:Sigma x X -> Sigma (not -> X)
and predicate a:X -> Sigma
f(a) & a = f(true) & a
where the parameter x has been suppressed from the equation
so in full it reads, less clearly
f(a(x),x) & a(x) = f(true,x) & a(x)
Paul
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