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author | Paul Phillips <paulp@improving.org> | 2012-09-27 06:41:17 -0700 |
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committer | Paul Phillips <paulp@improving.org> | 2012-09-27 09:24:04 -0700 |
commit | 96d4a8646b1962fac2f2fc443b56c6619221b43c (patch) | |
tree | bd39f411275452bb5db2335da966ac8042cd504f /src/library-aux | |
parent | 17b409b7832f541e3d52d2776c8ff3c47574ae0f (diff) | |
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Nailed down the "impossible match" logic.
I will again defer to a comment.
/** Given classes A and B, can it be shown that nothing which is
* an A will ever be a subclass of something which is a B? This
* entails not only showing that !(A isSubClass B) but that the
* same is true of all their subclasses. Restated for symmetry:
* the same value cannot be a member of both A and B.
*
* 1) A must not be a subclass of B, nor B of A (the trivial check)
* 2) One of A or B must be completely knowable (see isKnowable)
* 3) Assuming A is knowable, the proposition is true if
* !(A' isSubClass B) for all A', where A' is a subclass of A.
*
* Due to symmetry, the last condition applies as well in reverse.
*/
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