I was mistaken, the proof of [data:c e:in [data:a e:list [data:b e:list [data:c e:list e:empty]]]] is actually [e:ok e:and [e:ok e:and [e:ok e:and [data:c e:in [data:c e:list e:empty]]] e:thus [data:c e:in [data:b e:list [data:c e:list e:empty]]]] e:thus [data:c e:in [data:a e:list [data:b e:list [data:c e:list e:empty]]]]] -- Jos De Roo -- AGFAReceived on Friday, 11 August 2000 06:34:40 GMT
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