a common point,' 'having a separate point.' Hence if we take these two classes together, we include any Pair that can be proposed. Thus we get the Theorem 'Any Pair of Lines, which are separational from a third Line, are either coincidental or separational'; the predicate of which is equivalent to 'are not intersectional.'
Min. I see. And how are 16 (c) and 16 (b) related to 15?
Euc. Each of them is a Contranominal of 15 (a); and they are also contranominal to each other.
Min. I should like to see that drawn out.
Euc. Let 'A, B, C' be three Lines. Then 16 (a) may be written 'A Pair of Lines (A, B), which are separational from a third Line (C), are not intersectional with each other.'
This yields three Contranominals. The first is 'If A, B, are intersectional; it cannot be true that B, C, are separational, and also A, C.' i.e. 'A Pair of intersectional Lines (A, B) cannot both be separational from a third Line (C)': the second is 'If B, C are separational, and A, B intersectional; then A, C are not separational.' i.e. 'A Line (A) which is intersectional with one (B) of two separational Lines (B, C), is not separational from the other (C)': and the third proves a similar Theorem for B.
Min. Yes, but your conclusion now is 'A is not separational from C': whereas 16 (b) says 'is intersectional.'
Euc. That is so: but since A is intersectional with a Line (B) which is separational from C, it is axiomatic that it has a point separate from C, and so cannot be coincidental with it. Hence, its being 'not separational from C' proves that it must be intersectional with it.
