Re: Why is there something rather than nothing? From quantum theory to dialectics?
On 08 Feb 2015, at 13:30, spudboy100 via Everything List wrote: Bruno, are you familiar with the atheistic (so-called) theologies of Dr. Eric Steinhart? He's a bright philosopher from William Patterson University, is the US. He was originally a software engineer and is like yourself, a math guy. He applies his experience to his philosophy, and after reading your writings here, as well as Amoeba, his insights seem to parallel yours. Also, Clement Vidal's, as well. Every heard of him? His papers focus on the origins of the universe(s) Platonism, Computationalism, and Digital Philosophy. It's not exactly like your work, but it certainly parallels it. Ever heard of him? It sort of informs this topic I think. I don't think I know him although the name invke some familiarity. Did he got the first person indeterminacy, the mathematicalism or arithmeticallism? The mean to test this. You might sum up the idea, if you have the time, The problem with many scientists is that they stop doing science when doing philosophy. It is not a problem, but it can be confusing in that field. Bruno -Original Message- From: Samiya Illias samiyaill...@gmail.com To: everything-list everything-list@googlegroups.com Sent: Sat, Feb 7, 2015 11:07 pm Subject: Re: Why is there something rather than nothing? From quantum theory to dialectics? On Thu, Feb 5, 2015 at 8:27 PM, Bruno Marchal marc...@ulb.ac.be wrote: On 04 Feb 2015, at 17:14, Samiya Illias wrote: On Wed, Feb 4, 2015 at 5:49 PM, Bruno Marchal marc...@ulb.ac.be wrote: On 04 Feb 2015, at 06:02, Samiya Illias wrote: On 04-Feb-2015, at 12:01 am, Bruno Marchal marc...@ulb.ac.be wrote: Then reason shows that arithmetic is already full of life, indeed full of an infinity of universal machines competing to provide your infinitely many relatively consistent continuations. Incompleteness imposes, at least formally, a soul (a first person), an observer (a first person plural), a god (an independent simple but deep truth) to any machine believing in the RA axioms together with enough induction axioms. I know you believe in them. The lexicon is p truthGod []p provable Intelligible (modal logic, G and G*) []p p the soul (modal logic, S4Grz) []p t intelligible matter(with p sigma_1) (modal logic, Z1, Z1*) []p sensible matter (with p sigma_1) (modal logic, X1, X1*) You need to study some math, I have been wanting to but it seems such an uphill task. Yet, its a mountain I would like to climb :) 7 + 0 = 7. You are OK with this? Tell me. OK Are you OK with the generalisation? For all numbers n, n + 0 = n. Right? Right :) You suggest I begin with Set Theory? No need of set theory, as I have never been able to really prefer one theory or another. It is too much powerful, not fundamental. At some point naive set theory will be used, but just for making thing easier: it will never be part of the fundamental assumptions. I use only elementary arithmetic, so you need only to understand the following statements (and some other later): Please see if my assumptions/interpretations below are correct: x + 0 = x if x=1, then 1+0=1 x + successor(y) = successor(x + y) 1 + 2 = (1+2) = 3 I agree, but you don't show the use of the axiom: x + successor(y) = successor(x + y), or x +s(y) = s(x + y). I didn't use the axioms. I just substituted the axioms variables with the natural numbers. Are you OK? To avoid notational difficulties, I represent the numbers by their degree of parenthood (so to speak) with 0. Abbreviating s for successor: 0, s(0), s(s(0)), s(s(s(0))), ... If the sequence represents 0, 1, 2, 3, ... We can use 0, 1, 2, 3, ... as abbreviation for 0, s(0), s(s(0)), s(s(s(0))), ... Can you derive that s(s(0)) + s(0) = s(s(s(0))) with the statements just above? then 2 + 1 = 3 Hmm... s(s(0)) + s(0) = s(s(s(0))) is another writing for 2 + 1 = 3, but it is not clear if you proved it using the two axioms: 1) x + 0 = x 2) x + s(y)) = s(x + y) Let me show you: We must compute: s(s(0)) + s(0) The axiom 2) says that x + s(y) = s(x + y), for all x and y. We see that s(s(0)) + s(0) matches x + s(y), with x = s(s(0)), and y = 0. OK? So we can apply the axiom 2, and we get, by replacing x (= s(s(0))) and y (= 0) in the axiom 2). This gives s(s(0)) + s(0) = s( s(s(0)) + 0 ) OK? (this is a simple substitution, suggested by the axiom 2) But then by axiom 1, we know that s(s(0)) + 0 = s(s(0)), so the right side becomes s( s(s(0)) +0 ) = s( s(s(0)) ) So we have proved s(s(0)) + s(0) = s(s(s(0))) OK? Yes, thanks! Can you guess how many times you need to use the axiom 2) in case I would ask you to prove 1 + 8 = 9. You might do it for training purpose. 1+8=9 Translating in successor terms: s(0) + s(s(s(s(s(s(s(s(0 = s(s(s(s(s(s(s(s(s(0) Applying Axiom 2 by substituting x=8 or
Re: Why is there something rather than nothing? From quantum theory to dialectics?
On Tue, Feb 10, 2015 at 12:50 AM, Bruno Marchal marc...@ulb.ac.be wrote: On 08 Feb 2015, at 05:07, Samiya Illias wrote: On Thu, Feb 5, 2015 at 8:27 PM, Bruno Marchal marc...@ulb.ac.be wrote: On 04 Feb 2015, at 17:14, Samiya Illias wrote: On Wed, Feb 4, 2015 at 5:49 PM, Bruno Marchal marc...@ulb.ac.be wrote: On 04 Feb 2015, at 06:02, Samiya Illias wrote: On 04-Feb-2015, at 12:01 am, Bruno Marchal marc...@ulb.ac.be wrote: Then reason shows that arithmetic is already full of life, indeed full of an infinity of universal machines competing to provide your infinitely many relatively consistent continuations. Incompleteness imposes, at least formally, a soul (a first person), an observer (a first person plural), a god (an independent simple but deep truth) to any machine believing in the RA axioms together with enough induction axioms. I know you believe in them. The lexicon is p truthGod []p provable Intelligible (modal logic, G and G*) []p p the soul (modal logic, S4Grz) []p t intelligible matter(with p sigma_1) (modal logic, Z1, Z1*) []p sensible matter (with p sigma_1) (modal logic, X1, X1*) You need to study some math, I have been wanting to but it seems such an uphill task. Yet, its a mountain I would like to climb :) 7 + 0 = 7. You are OK with this? Tell me. OK Are you OK with the generalisation? For all numbers n, n + 0 = n. Right? Right :) You suggest I begin with Set Theory? No need of set theory, as I have never been able to really prefer one theory or another. It is too much powerful, not fundamental. At some point naive set theory will be used, but just for making thing easier: it will never be part of the fundamental assumptions. I use only elementary arithmetic, so you need only to understand the following statements (and some other later): Please see if my assumptions/interpretations below are correct: x + 0 = x if x=1, then 1+0=1 x + successor(y) = successor(x + y) 1 + 2 = (1+2) = 3 I agree, but you don't show the use of the axiom: x + successor(y) = successor(x + y), or x +s(y) = s(x + y). I didn't use the axioms. I just substituted the axioms variables with the natural numbers. And use your common intuition. Good. The idea now will be to see if the axioms given capture that intuition, fully, or in part. Are you OK? To avoid notational difficulties, I represent the numbers by their degree of parenthood (so to speak) with 0. Abbreviating s for successor: 0, s(0), s(s(0)), s(s(s(0))), ... If the sequence represents 0, 1, 2, 3, ... We can use 0, 1, 2, 3, ... as abbreviation for 0, s(0), s(s(0)), s(s(s(0))), ... Can you derive that s(s(0)) + s(0) = s(s(s(0))) with the statements just above? then 2 + 1 = 3 Hmm... s(s(0)) + s(0) = s(s(s(0))) is another writing for 2 + 1 = 3, but it is not clear if you proved it using the two axioms: 1) x + 0 = x 2) x + s(y)) = s(x + y) Let me show you: We must compute: s(s(0)) + s(0) The axiom 2) says that x + s(y) = s(x + y), for all x and y. We see that s(s(0)) + s(0) matches x + s(y), with x = s(s(0)), and y = 0. OK? So we can apply the axiom 2, and we get, by replacing x (= s(s(0))) and y (= 0) in the axiom 2). This gives s(s(0)) + s(0) = s( s(s(0)) + 0 ) OK? (this is a simple substitution, suggested by the axiom 2) But then by axiom 1, we know that s(s(0)) + 0 = s(s(0)), so the right side becomes s( s(s(0)) +0 ) = s( s(s(0)) ) So we have proved s(s(0)) + s(0) = s(s(s(0))) OK? Yes, thanks! You are welcome. Can you guess how many times you need to use the axiom 2) in case I would ask you to prove 1 + 8 = 9. You might do it for training purpose. 1+8=9 Translating in successor terms: s(0) + s(s(s(s(s(s(s(s(0 = s(s(s(s(s(s(s(s(s(0) Applying Axiom 2 by substituting x=8 or s(s(s(s(s(s(s(s(0, and y=0, s(s(s(s(s(s(s(s(0 + s(0) = s( s(s(s(s(s(s(s(s(0 + 0) Applying axiom 1 to the right side: s(0) + s(s(s(s(s(s(s(s(0 = s(s(s(s(s(s(s(s(s(0) 1+8=9 Is the above the correct method to arrive at the proof? I only used axiom 2 once. Am I missing some basic point? Let me see. Axiom 2 says:x + s(y)) = s(x + y). Well, if x = 8, and y = 0, we get 8 + 1, and your computation/proofs is correct, in that case. So you would have been correct if I was asking you to prove/compute that 8 + 1 = 9. Unfortunately I asked to prove/compute that 1 + 8 = 9. I think that you have (consciously?) use the fact that 1 + 8 = 8 + 1, which speeds the computation. Well, later I ill show you that the idea that for all x and y x + y = y + x, is NOT provable with the axioms given (despite that theorey will be shown to be already Turing Universal. No worry. Your move was clever, but you need to put yourself in the mind of a very stupid machine which understand only the axioms given. I understand Can you show that 1 + 8 = 9.
evangelizing robots
In two senses of that term! Or something. http://bigthink.com/ideafeed/robot-religion-2 http://gizmodo.com/when-superintelligent-ai-arrives-will-religions-try-t-1682837922 -- You received this message because you are subscribed to the Google Groups Everything List group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
Re: Why is there something rather than nothing? From quantum theory to dialectics?
From Wikipedia I get the idea that he is interested in the technological singularity, mind uploading and suchlike. On 10 February 2015 at 09:16, Bruno Marchal marc...@ulb.ac.be wrote: On 08 Feb 2015, at 13:30, spudboy100 via Everything List wrote: Bruno, are you familiar with the atheistic (so-called) theologies of Dr. Eric Steinhart? He's a bright philosopher from William Patterson University, is the US. He was originally a software engineer and is like yourself, a math guy. He applies his experience to his philosophy, and after reading your writings here, as well as Amoeba, his insights seem to parallel yours. Also, Clement Vidal's, as well. Every heard of him? His papers focus on the origins of the universe(s) Platonism, Computationalism, and Digital Philosophy. It's not exactly like your work, but it certainly parallels it. Ever heard of him? It sort of informs this topic I think. I don't think I know him although the name invke some familiarity. Did he got the first person indeterminacy, the mathematicalism or arithmeticallism? The mean to test this. You might sum up the idea, if you have the time, The problem with many scientists is that they stop doing science when doing philosophy. It is not a problem, but it can be confusing in that field. Bruno -Original Message- From: Samiya Illias samiyaill...@gmail.com To: everything-list everything-list@googlegroups.com Sent: Sat, Feb 7, 2015 11:07 pm Subject: Re: Why is there something rather than nothing? From quantum theory to dialectics? On Thu, Feb 5, 2015 at 8:27 PM, Bruno Marchal marc...@ulb.ac.be wrote: On 04 Feb 2015, at 17:14, Samiya Illias wrote: On Wed, Feb 4, 2015 at 5:49 PM, Bruno Marchal marc...@ulb.ac.be wrote: On 04 Feb 2015, at 06:02, Samiya Illias wrote: On 04-Feb-2015, at 12:01 am, Bruno Marchal marc...@ulb.ac.be wrote: Then reason shows that arithmetic is already full of life, indeed full of an infinity of universal machines competing to provide your infinitely many relatively consistent continuations. Incompleteness imposes, at least formally, a soul (a first person), an observer (a first person plural), a god (an independent simple but deep truth) to any machine believing in the RA axioms together with enough induction axioms. I know you believe in them. The lexicon is p truthGod []p provable Intelligible (modal logic, G and G*) []p p the soul (modal logic, S4Grz) []p t intelligible matter(with p sigma_1) (modal logic, Z1, Z1*) []p sensible matter (with p sigma_1) (modal logic, X1, X1*) You need to study some math, I have been wanting to but it seems such an uphill task. Yet, its a mountain I would like to climb :) 7 + 0 = 7. You are OK with this? Tell me. OK Are you OK with the generalisation? For all numbers n, n + 0 = n. Right? Right :) You suggest I begin with Set Theory? No need of set theory, as I have never been able to really prefer one theory or another. It is too much powerful, not fundamental. At some point naive set theory will be used, but just for making thing easier: it will never be part of the fundamental assumptions. I use only elementary arithmetic, so you need only to understand the following statements (and some other later): Please see if my assumptions/interpretations below are correct: x + 0 = x if x=1, then 1+0=1 x + successor(y) = successor(x + y) 1 + 2 = (1+2) = 3 I agree, but you don't show the use of the axiom: x + successor(y) = successor(x + y), or x +s(y) = s(x + y). I didn't use the axioms. I just substituted the axioms variables with the natural numbers. Are you OK? To avoid notational difficulties, I represent the numbers by their degree of parenthood (so to speak) with 0. Abbreviating s for successor: 0, s(0), s(s(0)), s(s(s(0))), ... If the sequence represents 0, 1, 2, 3, ... We can use 0, 1, 2, 3, ... as abbreviation for 0, s(0), s(s(0)), s(s(s(0))), ... Can you derive that s(s(0)) + s(0) = s(s(s(0))) with the statements just above? then 2 + 1 = 3 Hmm... s(s(0)) + s(0) = s(s(s(0))) is another writing for 2 + 1 = 3, but it is not clear if you proved it using the two axioms: 1) x + 0 = x 2) x + s(y)) = s(x + y) Let me show you: We must compute: s(s(0)) + s(0) The axiom 2) says that x + s(y) = s(x + y), for all x and y. We see that s(s(0)) + s(0) matches x + s(y), with x = s(s(0)), and y = 0. OK? So we can apply the axiom 2, and we get, by replacing x (= s(s(0))) and y (= 0) in the axiom 2). This gives s(s(0)) + s(0) = s( s(s(0)) + 0 ) OK? (this is a simple substitution, suggested by the axiom 2) But then by axiom 1, we know that s(s(0)) + 0 = s(s(0)), so the right side becomes s( s(s(0)) +0 ) = s( s(s(0)) ) So we have proved s(s(0)) + s(0) = s(s(s(0))) OK? Yes, thanks! Can you guess how many times you need to use the axiom 2) in case I would
Re: Why is there something rather than nothing? From quantum theory to dialectics?
On Mon, Feb 9, 2015 at 9:16 PM, Bruno Marchal marc...@ulb.ac.be wrote: On 08 Feb 2015, at 13:30, spudboy100 via Everything List wrote: Bruno, are you familiar with the atheistic (so-called) theologies of Dr. Eric Steinhart? He's a bright philosopher from William Patterson University, is the US. He was originally a software engineer and is like yourself, a math guy. He applies his experience to his philosophy, and after reading your writings here, as well as Amoeba, his insights seem to parallel yours. Also, *Clement Vidal*'s, as well. Every heard of him? His papers focus on the origins of the universe(s) Platonism, Computationalism, and Digital Philosophy. It's not exactly like your work, but it certainly parallels it. Ever heard of him? It sort of informs this topic I think. I don't think I know him although the name invke some familiarity. Did he got the first person indeterminacy, the mathematicalism or arithmeticallism? The mean to test this. You might sum up the idea, if you have the time, The problem with many scientists is that they stop doing science when doing philosophy. It is not a problem, but it can be confusing in that field. One of those names at least is familiar to the list because: On Fri, Oct 3, 2014 at 12:25 AM, spudboy100 via Everything List everything-list@googlegroups.com wrote: To die for Allah is slay for Allah. The reward for the mujahedeen is enormous, for to sacrifice ones self, and the opponent of God, is to granted immediate entry into paradise (Janah) and its rewards are not unsubstantial. One way to change the Umah's mind (if such is even possible) would be to make widespread, *Clement Vidal*'s publications, especially, The Beginning and the End, The Meaning of Life in a Cosmological Perspective. Part of the book details with afterlife concepts in a rational sense, as well as much, more. *Vidal* is a colleague of Bruno Marchal at Free University, Brussels. * Vidal*'s influence may induce those looking for a heavenly, rewards, for head chopping unbelievers, a good think. It would also alter our own perspective as well. Make the world a bit more peaceful, and provide some reassurance for all. Honk! If you all agree ;-) On Fri, Oct 3, 2014 at 11:52 PM, spudboy100 via Everything List everything-list@googlegroups.com wrote: Bruno might comment on his colleague, at university, *Clement Vidal*. The Evo-Devo approach, etc. On Tue, Sep 30, 2014 at 3:46 AM, spudboy100 via Everything List everything-list@googlegroups.com wrote: Sent from AOL Mobile If We remain a sophisticated civilization, we should be able to leave space exploration for the robots, until such a time and and place where we uncover something dynamically interesting. But this is for another generation to decide, and not ours. Which makes things, seem, our world, our times, our worries, seem so temporary. On the other hand I am now reading the work by *Clement Vidal*, of Free University, Brussels, a colleague, of Bruno Marchal, on the meaning and purpose of intelligence, life, and cosmology, leading to the far future-a very, different perspective indeed. On Wed, Jun 25, 2014 at 5:55 PM, spudboy100 via Everything List everything-list@googlegroups.com wrote: Dr. Marchal, do you ever get in conversations with your fellow academician,* Clement Vidal*? He's a philosopher at your University? Do you ever get into the Evo-Devo view? So that may be part of the reason the name is slowly becoming familiar although I wouldn't know, as I can't search the list's archives completely, nor do I receive/want all posts archived in my Inbox, therefore filtering and/or ignoring a lot. PGC -- You received this message because you are subscribed to the Google Groups Everything List group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.
Re: Why is there something rather than nothing? From quantum theory to dialectics?
On 08 Feb 2015, at 05:07, Samiya Illias wrote: On Thu, Feb 5, 2015 at 8:27 PM, Bruno Marchal marc...@ulb.ac.be wrote: On 04 Feb 2015, at 17:14, Samiya Illias wrote: On Wed, Feb 4, 2015 at 5:49 PM, Bruno Marchal marc...@ulb.ac.be wrote: On 04 Feb 2015, at 06:02, Samiya Illias wrote: On 04-Feb-2015, at 12:01 am, Bruno Marchal marc...@ulb.ac.be wrote: Then reason shows that arithmetic is already full of life, indeed full of an infinity of universal machines competing to provide your infinitely many relatively consistent continuations. Incompleteness imposes, at least formally, a soul (a first person), an observer (a first person plural), a god (an independent simple but deep truth) to any machine believing in the RA axioms together with enough induction axioms. I know you believe in them. The lexicon is p truthGod []p provable Intelligible (modal logic, G and G*) []p p the soul (modal logic, S4Grz) []p t intelligible matter(with p sigma_1) (modal logic, Z1, Z1*) []p sensible matter (with p sigma_1) (modal logic, X1, X1*) You need to study some math, I have been wanting to but it seems such an uphill task. Yet, its a mountain I would like to climb :) 7 + 0 = 7. You are OK with this? Tell me. OK Are you OK with the generalisation? For all numbers n, n + 0 = n. Right? Right :) You suggest I begin with Set Theory? No need of set theory, as I have never been able to really prefer one theory or another. It is too much powerful, not fundamental. At some point naive set theory will be used, but just for making thing easier: it will never be part of the fundamental assumptions. I use only elementary arithmetic, so you need only to understand the following statements (and some other later): Please see if my assumptions/interpretations below are correct: x + 0 = x if x=1, then 1+0=1 x + successor(y) = successor(x + y) 1 + 2 = (1+2) = 3 I agree, but you don't show the use of the axiom: x + successor(y) = successor(x + y), or x +s(y) = s(x + y). I didn't use the axioms. I just substituted the axioms variables with the natural numbers. And use your common intuition. Good. The idea now will be to see if the axioms given capture that intuition, fully, or in part. Are you OK? To avoid notational difficulties, I represent the numbers by their degree of parenthood (so to speak) with 0. Abbreviating s for successor: 0, s(0), s(s(0)), s(s(s(0))), ... If the sequence represents 0, 1, 2, 3, ... We can use 0, 1, 2, 3, ... as abbreviation for 0, s(0), s(s(0)), s(s(s(0))), ... Can you derive that s(s(0)) + s(0) = s(s(s(0))) with the statements just above? then 2 + 1 = 3 Hmm... s(s(0)) + s(0) = s(s(s(0))) is another writing for 2 + 1 = 3, but it is not clear if you proved it using the two axioms: 1) x + 0 = x 2) x + s(y)) = s(x + y) Let me show you: We must compute: s(s(0)) + s(0) The axiom 2) says that x + s(y) = s(x + y), for all x and y. We see that s(s(0)) + s(0) matches x + s(y), with x = s(s(0)), and y = 0. OK? So we can apply the axiom 2, and we get, by replacing x (= s(s(0))) and y (= 0) in the axiom 2). This gives s(s(0)) + s(0) = s( s(s(0)) + 0 ) OK? (this is a simple substitution, suggested by the axiom 2) But then by axiom 1, we know that s(s(0)) + 0 = s(s(0)), so the right side becomes s( s(s(0)) +0 ) = s( s(s(0)) ) So we have proved s(s(0)) + s(0) = s(s(s(0))) OK? Yes, thanks! You are welcome. Can you guess how many times you need to use the axiom 2) in case I would ask you to prove 1 + 8 = 9. You might do it for training purpose. 1+8=9 Translating in successor terms: s(0) + s(s(s(s(s(s(s(s(0 = s(s(s(s(s(s(s(s(s(0) Applying Axiom 2 by substituting x=8 or s(s(s(s(s(s(s(s(0, and y=0, s(s(s(s(s(s(s(s(0 + s(0) = s( s(s(s(s(s(s(s(s(0 + 0) Applying axiom 1 to the right side: s(0) + s(s(s(s(s(s(s(s(0 = s(s(s(s(s(s(s(s(s(0) 1+8=9 Is the above the correct method to arrive at the proof? I only used axiom 2 once. Am I missing some basic point? Let me see. Axiom 2 says:x + s(y)) = s(x + y). Well, if x = 8, and y = 0, we get 8 + 1, and your computation/proofs is correct, in that case. So you would have been correct if I was asking you to prove/compute that 8 + 1 = 9. Unfortunately I asked to prove/compute that 1 + 8 = 9. I think that you have (consciously?) use the fact that 1 + 8 = 8 + 1, which speeds the computation. Well, later I ill show you that the idea that for all x and y x + y = y + x, is NOT provable with the axioms given (despite that theorey will be shown to be already Turing Universal. No worry. Your move was clever, but you need to put yourself in the mind of a very stupid machine which understand only the axioms given. Can you show that 1 + 8 = 9. Better, tell me how many times you will need to use the second axioms? Let me ask you
