George Hotz | Programming | twitchcoq : pt 2, can we prove true is not false | Coq (software)

Date of stream 2 Nov 2019. Live-stream chat added as Subtitles/CC – English (Twitch Chat). Stream title: twitchcoq : pt 2, can we prove true is not false Video …


  1. true implies false only for one time or once ever if we're talking about states and also iff false implies true after stating that true implies false.
    false will always come before true and if it were denoted by some distance but true and false and then state false <= true then it will always be that there is some canonical logarithmic singularity of chance that should show that there is a greater likely hood of whatever this state or thing is to be false than true if it were some deterministic quantized value that they wanted.

  2. 1:23:19 when it speaks about observing and not observing some specific scenario what it is referring to as far as I know is that it is speaking about observable and non observable operators acting on some functions such that if the difference of the order rearranged of two operators on some wave function is 0 then it means that the order we calculate what the out put value of what the operators individually on the wave function per say gives you and being able to calculate something from a given output or understanding the output without lets say imaginary numbers then it is known as observable meaning you can measure the output you got with an operator..,.sorry a better def is the hermitian conjugate where the order of this operator lets say the distance operator on a function will just give you the distance x * function and the order doesn't matter in a sense that some function f * x can be rearranged making it observable and a non observable thing is a momentum operator in physics in elementary particle physics which is denoted
    p = i(h_reduced_planck's_constant)d/dx obviously the 'd/dx' is the part that has to act on a function and if you did pf and then fp you realize that the pf gives you some value(f) but fp gives f(p) and it still has to act on something until you get an output value and even if you got an imaginary number out it would be considered also not observable because the idea that 3 meters * imaginary is an output doesnt mean anything to us but if you got 3 meters ok perfect observable and if you still get an output value where you still need to make f(p()) act on something then that is still considered and non obvservable…I know this is harder to explain but that's just literally the first fucking sentence sorry forgive me I tried to explain but if I code this it'll come out so pretty you'd be able to read it nicely.


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