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Re: NAND & NOR now n-ary; Bug in BITWISE-EQV? implementation
> I also discovered a bug in the spec for n-ary eqv. This is true:
> (eqv i j) = (not (xor i j))
> But it does not generalise to the n-ary case. That is, it is *NOT TRUE* that
> (eqv i ...) = (not (xor i ...))
Are you sure? I've always thought that (eqv i ...) = (not (xor i ...)) .
What does eqv mean anyway? Does a 1 bit in the result mean that all arguments
have the same value for that bit? Or does it mean even-parity (odd-parity
being computed by xor)? What's your reference?