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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?