 ThunderCats -2x5- Thundercats Ho (V) Pt 2
Descrição do Vídeo:
Hachiman pulls Lion-O onto the bridge sensing that
Lion-O is not a dishonorable foe. Hachiman realizes that
Mumm-Ra ticked him and vows to help Lion-O free Lynx-O,
Ben-Gali and Pumyra.
Miles away, Cheetara and Tygra are making their way to
the rendezvous point when Captain Shiner attack them at
the demand of Mumm-Ra. The two cats work together to send
off Shiner's ship into deep space before continuing on.
Nearby on the sea, the Gomplin can no longer tow the
Berserkers' ship and Turmagar cuts the line. As the
pirates regain their ship, Turmagar and the Gomplin
decide to attack again. Bu
Data: November 22, 2008
Palavras Relacionados: none
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Final Fantasy X - NSGNSND Monster Arena - Nega Elemental 1
Descrição do Vídeo:
Video of Nega Elemental being defeated in NSGNSND (No
Sphere Grid, No Summoning, No Doom) challenge.
NOTE: I have made exceptions to the No Sphere Grid rule
for "Flee" and "Bribe". The former is to quickly escape
from unwanted battles and has no
effect on any battle shown. "Bribe" is used to gather
items quickly, and likewise, has no effect on any battle
shown.
Equipment:
All characters weapons: First Strike, Evade & Counter,
Alchemy, Triple Overdrive (only First Strike is relevant
in this battle, you might want to chuck on Magic Counter
so you can throw Tetra Elementals around and ge
Data: November 22, 2008
Palavras Relacionados: final fantasy nsgnsnd monster arena nega elemental
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WVOQ 51 Willard Van Orman Quine - cleancut
Descrição do Vídeo:
http://www-groups.dcs.st-and.ac.uk/~history/Printonly
/Quine.html - a ? fold for nats primrec iter g a 0 = a
iter g a (Suc n) = g (iter g a n) So iterating the step
function n times on an initial sequent s gives the
sequentsat depth n in the derivation. lemma ? x. ((n,x) ?
deriv s) = (x ? set (iter step [s] n)) Now we know that
the derivation from s is finite iff there is some n
such that there are no sequents at depth n in the
derivation, in which case the nth iteration of the step
function on s will be empty. lemma finite-deriv: finite
(deriv s) = (? m. iter step [s] m = []) No
Data: November 22, 2008
Palavras Relacionados: Philosophy Willard Van Orman Quine
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WVOQ 48 Willard Van Orman Quine - cleancut
Descrição do Vídeo:
http://www-groups.dcs.st-and.ac.uk/~history/Printonly
/Quine.html - ] =? ¬ feval (model (ns-of-s s)) ntou A Let
us treat the case that (?x.P x) n appears on f. Then we
know that (?x.P x) 0 appears on f. Then (?x.P x) 0
eventually gets considered, and P x 0 and (?x.P x) 1
appears on f. Then (?x.P x) 1 eventually gets
considered, and P x 1 and (?x.P x) 2 appears on f.
Continuing in this way, we see that for all n, P x n
appears on f. Applying the induction hypothesis to each
of these, we see that for all n, the interpretation of P
x n , i.e. P n, is false in the model. So "t
Data: November 22, 2008
Palavras Relacionados: Philosophy Willard Van Orman Quine
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WVOQ 46 Willard Van Orman Quine - cleancut
Descrição do Vídeo:
http://www-groups.dcs.st-and.ac.uk/~history/Printonly
/Quine.html - the notion of when a formula is
considered at a point n in a path, which is when the
formula is at the head of the sequent at position n.
constdefs contains :: (nat ? (nat × nseq)) ? nat ?
nform ? bool contains f n nf ? nf ? set (snd (f n))
constdefs considers :: (nat ? (nat × nseq)) ? nat ?
nform ? bool considers f n nf ? case snd (f n) of [] ?
False | (x#xs) ? x = nf 3.8 Models 2 A falsifying model
will in general consist of at least countably many
elements. So far, we have said nothing about the size
of
Data: November 22, 2008
Palavras Relacionados: Philosophy Willard Van Orman Quine
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WVOQ 42 Willard Van Orman Quine - cleancut
Descrição do Vídeo:
http://www-groups.dcs.st-and.ac.uk/~history/Printonly
/Quine.html - case of the ? rule, which we prove for
arbitrary fresh u. This case depends on the following
lemma, which states roughly that evaluating [u/x]f in
environment e is equivalent to evaluating f in an
environment e(x := e u), i.e. in the same environment
except that the free variable x gets mapped to whatever
u was mapped to by the original environment e. lemma
feval-finst: feval MI e (finst A u) = feval MI
(nat-case (e u) e) A The statement of the main lemma is
as follows. lemma sound-FAll: u /? set (sfv (FAll f #s
Data: November 22, 2008
Palavras Relacionados: Philosophy Willard Van Orman Quine
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WVOQ 40 Willard Van Orman Quine - cleancut
Descrição do Vídeo:
http://www-groups.dcs.st-and.ac.uk/~history/Printonly
/Quine.html - the truth of a formula, using standard
Tarski semantics. consts feval :: model ? env ? form ?
bool 3 This is one of only two places where we make
axiomatic assertions. Both could be avoided by using
existing type ind instead of declaring U. Page 8 8
primrec feval MI e (PAtom P vs) = (let IP = (snd MI ) P
in IP (map e vs)) feval MI e (NAtom P vs) = (let IP =
(snd MI ) P in ¬ (IP (map e vs))) feval MI e (FConj f g)
= ((feval MI e f ) ? (feval MI e g)) feval MI e (FDisj f
g) = ((feval MI e f ) ? (feval MI e g)) fe
Data: November 22, 2008
Palavras Relacionados: Philosophy Willard Van Orman Quine
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WVOQ 37 Willard Van Orman Quine - cleancut
Descrição do Vídeo:
http://www-groups.dcs.st-and.ac.uk/~history/Printonly
/Quine.html - that the proofs in the rest of the
mechanisation are uniform. ? Nil ? Then to represent
the rules, we use a function mapping the conclusion of
arule to a list of premises. consts subs :: nseq ? nseq
list primrec subs [] = [[]] subs (x#xs) = Page 7 7
(let (m,f ) = x in case f of PAtom p vs ? if NAtom p vs
? set (map snd xs) then [] else [xs@[(0,PAtom p vs)]] |
NAtom p vs ? if (PAtom p vs) ? set (map snd xs) then []
else [xs@[(0,NAtom p vs)]] | FConj f g ? [xs@[(0,f
)],xs@[(0,g)]] | FDisj f g ? [xs@[(0,f ),(0,
Data: November 22, 2008
Palavras Relacionados: Philosophy Willard Van Orman Quine
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WVOQ 28 Willard Van Orman Quine - cleancut
Descrição do Vídeo:
http://www-groups.dcs.st-and.ac.uk/~history/Printonly
/Quine.html - i as the number i, and interpreting an
atomic predicate P(x i 1 , . . . , x i n ) as true
iff P(x i 1 , . . . , x i n ) does not appear
anywhere on the failing path f. By induction on the size
of the formula, we can see that every formula appearing
on f must get false. Since f starts at s, and all
formulae in s get false, then s must also get false in
this model. Let us consider the ? case in the inductive
argument. We must show that if P ? Q appears on f, then
P ? Q gets false in the model. Assume P
Data: November 22, 2008
Palavras Relacionados: Philosophy Willard Van Orman Quine
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BED AND BREAKFAST
Descrição do Vídeo:
Mark Theodossiou plays a new cop in a small country
town who bites off more than he can chew in order to
crack a case. Starring Penny Brown-O'Dare, Paul Bryant
and John Tiplady. Produced and directed by Dorothy
Hartnett, 2008.
Data: November 22, 2008
Palavras Relacionados: Mark Theodossiou short film drama performing arts entertainment
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