This article was not illogic enough! Add more illogic!.
One hundred years ago (plus or minus a century or so) The most prominent heretics of the church of logic, the real masters of Swedish Al-Qaeda , came together to discuss their sinful logic and its implications. These masters of black magic included but were not limited to minds like Albert Newton, Issac Einstein, Archimedes von Tesla, and the nefarious ConfusUs the third. However, because the word logic has only 5 letters, it was not divisible by two. This resulted in the great logic schism, where the traditionalist "Logs" forfeited their lives for an eternity of floating down rivers, while the more radical IC's opted for the unoriginal yet tried and true method of just bombing the shit out of everyone.
For every number n and every formula F(y), where y is a free variable, we define q(n,G(F)), a relation between two numbers n and G(F), such that it corresponds to the statement "n is not the Gödel number of a proof of F(G(F))". Here, F(G(F)) can be understood as F with its own Gödel number as its argument.
Note that q takes as an argument G(F). the Gödel number of F. In order to prove either q(n,G(F)), or q(n,G(F)), it is necessary to perform number-theoretic operations on G(F) that mirror the following steps: decode the number G(F) into the formula F, replace all occurrences of y in F with the number G(F), and then compute the Gödel number of the resulting formula F(G(F)).
Note that for every specific number n and formula F(y), q(n,G(F)) is a straightforward (though complicated) arithmetical relation between two numbers n and G(F), building on the relation PF defined earlier. Further, q(n,G(F)) is provable if the finite list of formulas encoded by n is not a proof of F(G(F)), and q(n,G(F)) is provable if the finite list of formulas encoded by n is a proof of F(G(F)). Given any numbers n and G(F), either q(n,G(F)) or q(n,G(F)) (but not both) is provable.
Any proof of F(G(F)) can be encoded by a Gödel number n, such that q(n,G(F)) does not hold. If q(n,G(F)) holds for all natural numbers n, then there is no proof of F(G(F)). In other words, , a formula about natural numbers, corresponds to "there is no proof of F(G(F))".
We now define the formula P(x)=, where x is a free variable. The formula P itself has a Gödel number G(P) as does every formula.
This formula has a free variable x. Suppose we replace it with G(F), the Gödel number of a formula F(z), where z is a free variable. Then, P(G(F))= corresponds to "there is no proof of F(G(F))", as we have seen.
Consider the formula P(G(P))=. This formula concerning the number G(P) corresponds to "there is no proof of P(G(P))". We have here the self-referential feature that is crucial to the proof: A formula of the formal theory that somehow relates to its own provability within that formal theory. Very informally, P(G(P)) says: "I am not provable".
We will now show that neither the formula P(G(P)), nor its negation P(G(P), is provable.
Suppose P(G(P))= is provable. Let n be the Gödel number of a proof of P(G(P)). Then, as seen earlier, the formula q(n,G(P)) is provable. Proving both
q(n,G(P)) and violates the consistency of the formal theory. We therefore conclude that P(G(P) is not provable.
Consider any number n. Suppose q(n,G(P)) is provable. Then, n must be the Gödel number of a proof of P(G(P)). But we have just proved that P(G(P) is not provable. Since either q(n,G(P)) or q(n,G(P)) must be provable, we conclude that, for all natural numbers n, q(n,G(P)) is provable.
Suppose the negation of P(G(P)), P(G(P)) = , is provable. Proving both , and q(n,G(P)), for all natural numbers n, violates ω-consistency of the formal theory. Thus if the theory is ω-consistent, P(G(P)) is not provable.
We have sketched a proof showing that:
For any formal, recursively enumerable (i.e. effectively generated) theory of Peano Arithmetic,
- if it is consistent, then there exists an unprovable formula (in the language of that theory).
- if it is ω-consistent, then there exists a formula such that both it and its negation are unprovable.
Anything = Anything (substituting Anything for A)
Pineapple = Orange (This step relies on the fact that Pineapple, Orange, etc. are subtypes of Anything.)
Proof 2.5 (two and a pint of fives)
Proof that Alexander the Great's Horse was Not a Mammal
- Alexander the Great rode a "horse" (vis. assorted history tomes)
- In the habit of horses everywhere, on the front end, Alexander the Great's horse had forelegs.
- On the back end, his horse had two more legs (referred to by historians as "hind legs").
- Four legs (on the front) plus two more legs (on the back) make six legs.
- Six is an even number.
- Six is a most odd number of legs for a horse, as I'm sure you'll agree.
- Therefore Alexander the Great's horse had a number of legs which was both even and odd.
- The only number that is both even and odd is infinity.
- Therefore Alexander the Great's horse had an infinite number of legs.
- But infinity is larger than the number of legs on any mammal (certainly) and any insect (for sure) and, in fact, if you name any animal which is not a centipede, you'll find it has fewer than an infinite number of legs.
- Therefore, Alexander the Great, in reality, rode a centipede, not a horse at all.
- And a centipede is most certainly not a mammal.
- Q. E. D.
Logic is self defeating!!! They say that the wonders of Circular Arguments is evil!!! However, they do it too!!!
Premise #1: Logic exists. Premise #2: I know this because it's logical. Conclusion: Therefore, Logic exists.
You see??? They have to use circular logic to prove logic!!! The very thing that evil Logicans hate!!!
Given that cos4(∑am+b) for all a < b ≤ a+b is congruent to the fourth dimensional representation of the hyper-spherical analog to a hypercube ABCDE, express, in simplest form, the three-dimensional representation of a cubic analog of a sphere R.
This problem is quite simple when broken down into 3 simple higher-dimensional steps.
- Step 1: The equations rendering a hypercube are, of course,
z = n, y = n, x = n, and w = n, with n being an arbitrary number, and living inside set S. The hyper-spherical analog of set S, then, is simply the spherical analog of each equation, or w4+x4+y4+z4 = n, returning a hyper-sphere with radius n, and with points inside S'.
- Step 2: Now we have two equation that are both modeling this hyper-cube analog. Therefore, they are equal to each other and we get
cos4(∑am+b) = w4+x4+y4+z4 = n for all a < b ≤ a+b. By using our cosine rules and summation laws, the equation is simplified down to sin3(a+b) - cos2(a-b) + tan(an/bn) - n + 1/4w3-4x3+1/4y3-4z3) = 0, which is a valid spherical analog of a cube.
- Step 3: Finally, the spherical analog of a cube must be subverted into a cubic analog of a sphere. Therefore, using Sphere Subversion Laws II and VIII, one gets the equations
x = -1/4n3+a, y = 4n3-b, z = -n3/m.
Use as a trash can
Proponents of Logic (heretics and liars)
- The Vulcan species
- 19th Century German philosophers who talk to horses when they've grown old and senile
- Some Greek guy who yammered about caves and shadows
- A guy who wrote a book so dense and impenetrable, he had to write another book on how to understand it
- your older brother, who never could develop an effective comeback to "NANANANANA I'm not listening!", despite having been an honors student. SCREW YOU STEVE! MY REAL BROTHER IS DEAD!
Notes of the Feet
- which means, as any grad student can tell you, 'Quite Elegantly Demonstrated'
- This story comes from an old vague memory, so it may not be true. As I remember it, Arthur Schopenhauer became senile at the end of his life, and was seen discussing philosophy with horses
- Martin Heidegger wrote Being and Time, followed up by On Being and Time.