A neat identity I happened upon: the negation of an equivalence is equivalent to an exclusive or.
Equivalence: A <--> B
Or (A --> B) & (B --> A)
Or (~A v B) & (~B v A)
Negating this last
~[(~A v B) & (~B v A)]
and then applying DeMorgan's law, we get:
~(~A v B) v ~(~B v A)
Apply DeMorgan again to the parenthesized symbols for:
(A&~B) v (B&~A)
This last offers only mutually exclusive options.
|
|
Some Ryle essay footnotes appear on this site, which is devoted to life after physics.
There is no claim of expertise with respect to the musings on this page. Never use my stuff for homework!
Thursday, February 13, 2020
Trivial but neat
Subscribe to:
Post Comments (Atom)
Your example: [(p-->q) + (q --> p)] [~(p-->q) v (~q --> p]).~[q p] Dot means "and." The above can be r...
-
The well-ordering theorem, also known as Zermelo's theorem, says that every set can be well-ordered. A set X is well-ordered by a strict...
-
Claim: Every odd integer square is an element of some Pythagorean triple. We first show that every positive odd number 2n+1 is equivalent ...
-
Added Jan. 19, 2023 Footnote Xz.151. Freud in The Interpretation of Dreams (1914 translation by A.A. Brill) highlights the idea that the d...
No comments:
Post a Comment