Your FAQs cover situations where ALL the tests need to be simultaneously satisfied (known in set theory as intersection, and in Boolean algebra as the AND'ing of conditions).[ ] This is achieved by "multiplying" the logical tests:
[ ] [ ] [ ](test 1) * (test 2) * (test 3) * ...
You can also accommodate situations where only ONE OR MORE of the tests need to be satisfied (known in set theory as union, and in Boolean algebra as the OR'ing of conditions).[ ] You can do this by "adding" the logical tests, then using the SIGN function to reduce to 1 any resulting number that is greater than 1.[ ] Thus:
[ ] [ ] [ ]SIGN( (test 1) + (test 2) + (test 3) + ... )
You can then build up even more complicated compound tests, with unions embedded in intersections and vice versa.