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> 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? Marc