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Willem F Esterhuyse

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  1. It has a justification for chopping up sentences added. Its a no-brainer.
  2. So that you could see whether the higher thought functions is activated. The program leaves out quoted posts in the quoted post. We know what part of the brain registers conscious thought, if this part of the brain is activated at the same time as some higher thought area, we know it must have been a conscious thought.
  3. Here is a better version: Proof of "Axioms" of Propositional Logic: Synopsis. Willem F. Esterhuyse. Abstract. We introduce more basic axioms with which we are able to prove some "axioms" of Propositional Logic. We use the symbols from my other article: "Introduction to Logical Structures". Logical Structures (SrL) are graphs with doubly labelled vertices with edges carrying symbols. The proofs are very mechanical and does not require ingenuity to construct. It is easy to see that in order to transform information, it has to be chopped up. Just look at a kid playing with blocks with letters on them: he has to break up the word into letters to assemble another word. Within SrL we take as our "atoms" propositions with chopped up relations attached to them. We call the results: (incomplete) "structures". We play it safe by allowing only relations among propositions to be choppable. We will see whether this is the correct way of chopping up sentences (it seems to be). This is where our Attractors (Repulsors) and Stoppers come in. Attractors that face away from each other repels and so break a relation between the two propositions. Then a Stopper attaches to the chopped up relation to indicate it can't reconnect. So it is possible to infer sentences from sentences. The rules I stumbled upon, to implement this, seems to be consistent. 1. Introduction. We use new operators called "Attractors" and "Stoppers". An Attractor ( symbol: "-(" OR ")-") is an edge with a half circle symbol, that can carry any relation symbol. Axioms for Attractors include A:AA (Axiom: Attractor Annihilation) where we have as premise two structures named B with Attractors carrying the "therefore" symbol facing each other and attached to two neighboring structures: B. Because the structures are the same and the Attractors face each other, and the therefore symbols point in the same direction, they annihilate the structures B and we are left with a conclusion of the empty structure. Like in: ((B)->-( )->-(B)) <-> (Empty Structure). where "<->" means: "is equivalent to" or "follows from and vice versa". We also have the axiom: A:AtI (Attractor Introduction) in which we have a row of structures as premise and conclusion of the same row of structures each with an Attractor attached to them and pointing to the right or left. Like in: A B C D <-> (A)-( (B)-( (C)-( (D)-( OR: A B C D <-> )-(A) )-(B) )-(C) )-(D) A:AD distributes the Attractors and cut relations and places a Stopper on the chopped relation (see line 3 below). Stopper = "|-" or "-|". Further axioms are: A:SD says that we may drop a Stopper at either end of a line. And A:ASS says we can exchange Stoppers for Attractors (and vice versa) in a line of structures as long as we replace every instance of the operators. A:AL says we can link two attractors pointing towards each other and attached to two different structures. We can prove: P OR P -> P. We prove Modus Ponens as follows: Line nr. Statement Reason 1 B B -> C Premise 2 (B)->-( (B -> C)->-( 1, A:AtI 3 (B)->-( )->-(B) |->-(C)->-( 2, A:AD 4 |->-(C)->-( 3, A:AA 5 (C)->-( 4, A:SD 6 (C)->-| 5, A:ASS 7 C 6, A:SD We see that the Attractors cuts two structures into three (line 2 to line 3). We can prove AND-elimination, AND-introduction and transposition. We prove Theorem: AND introduction (T:ANDI): 1 A B Premise 2 A -(x)-( B -(x)-( 1, A:AtI 3 (A)-(x)-| (B)-(x)-| 2, A:ASS 4 (A)-(x)-| B 3, A:SD 5 (A)-(x)-( B 4, A:ASS 6 (A)-(x)-(B) 5, T:AL where "-(x)-" = "AND", and T:AL is a theorem to be proved by reasoning backwards through: 1 A -(x)- B Premise 2 A -(x)- B -(x)-( 1, A:AtI 3 )-(x)-(A) |-(x)-(B)-(x)-( 2, A:AD 4 |-(x)-(A) )-(x)-(B)-(x)-| 3, A:ASS 5 A )-(x)-(B) 4, A:SD. where the mirror image of this is proved similarly. Modus Tollens and Syllogism can also be proven with these axioms. We prove: Theorem (T:O): (A OR A) -> A: 1 A -(+)- A Premise 2 A -(+)- A -(+)-( 1, A:AtI 3 )-(+)-(A) |-(+)-(A)-(+)-( 2, A:AD 4 |-(+)-(A) )-(+)-(A)-(+)-| 3, A:ASS 5 A )-(+)-(A) 4, A:SDx2 6 A |-(+)-(A) 5, A:ASS and from this (on introduction of a model taking only structures with truth tables as real) we can conclude that A holds as required. We prove Syllogism: 1 A -> B B -> C Premise 2 (A -> B)->-( (B -> C)->-( 1, A:AtI 3 )->-(A)->-| (B)->-( )->-(B) |->-(C)->-( 2, A:ADx2 4 (A)->-| (B)->-( )->-(B) |->-(C) 3, A:ASS, A:SDx2, A:ASS 5 (A)->-| |->-(C) 4, A:AA 6 (A)->-( )->-(C) 5, A:ASS 7 A -> C 6, A:AL
  4. So that you could see whether the higher thought functions is activated. The program leaves out quoted posts in the quoted post.
  5. It is a test of awareness of perceptions and the act of perceiving.
  6. Yes I am sure. A good test would be to see if the person undergoing the test is proficient in working with words like: "listen" instead of "hear" and "see" instead of "look".
  7. I have seen a mathematical definition of consciousness somewhere online. Although I can't criticize it.
  8. Here is the Synopsis: Proof of "Axioms" of Propositional Logic: Synopsis. Willem F. Esterhuyse. Abstract We introduce more basic axioms with which we are able to prove some "axioms" of Propositional Logic. We use the symbols from my other article: "Introduction to Logical Structures". Logical Structures are graphs with doubly labelled vertices with edges carrying symbols. The proofs are very mechanical and does not require ingenuity to construct. We use new operators called "Attractors" and "Stoppers". An Attractor ( symbol: "-(" OR ")-") is an edge with a half circle symbol, that can carry any relation symbol. Axioms for Attractors include A:AA where we have as premise two structures named B with Attractors carrying the "therefore" symbol facing each other and attached to two neighbouring structures: B. Because the structures are the same and the Attractors face each other, and the therefore symbols point in the same direction they annihilate the structures B and we are left with a conclusion of the empty structure. Like in: ((B)->-( )->-(B)) -> (Empty Structure). We also have the axiom: A:AtI (Attractor Introduction) in which we have a row of structures as premise and conclusion of the same row of structures each with an Attractor attached to them and pointing to the right or left. Like in: A B C D -> (A)-( (B)-( (C)-( (D)-(. A:AD distributes the Attractors and cut relations and places a Stopper on the cutted relation (see line 3 below). Stopper = "|-". Further axioms are: A:SD says that we may drop a Stopper at either end of a line. And A:ASS says we can exchange Stoppers for Attractors in a line of structures as long as we replace every instance of the operators. We can prove: P OR P = P. We prove Modus Ponens as follows: Line nr. Statement Reason 1 B B -> C Premise 2 B ->-( (B -> C)->-( 1, A:AtI 3 B ->-( )->-(B) |->-(C)->-( 2, A:AD 4 |->-(C)->-( 3, A:AA 5 (C)->-( 4, A:SD 6 (C)->-| 5, A:ASS 7 C 6, A:SD We can prove AND-elimination, AND-introduction and transposition. We prove AND introduction (T:ANDI): 1 A B Premise 2 (A)-(x)-( (B) -(x)-( 1, A:AtI 3 (A)-(x)-| (B)-(x)-| 2, A:ASS 4 (A)-(x)-| B 3, A:SD 5 (A)-(x)-( B 4, A:ASS 6 (A)-(x)-(B) 5, T:AL where "-(x)-" = "AND", and T:AL is a theorem to be proved by reasoning backwards through: 1 A -(x)- B Premise 2 A -(x)- B -(x)-( 1, A:AtI 3 )-(x)-(A) |-(x)-(B)-(x)-( 2, A:AD 4 |-(x)-(A) )-(x)-(B)-(x)-| 3, A:ASS 5 A )-(x)-(B) 4, A:SD. where the mirror image of this is proved similarly. Modus Tollens and Syllogism can also be proven with these axioms. We prove: Theorem (T:O): (A OR A) -> A: 1 A -(+)- A Premise 2 A -(+)- A -(+)-( 1, A:AtI 3 )-(+)-(A) |-(+)-(A)-(+)-( 2, A:AD 4 |-(+)-(A) )-(+)-(A)-(+)-| 3, A:ASS 5 A )-(+)-(A) 4, A:SD 6 A |-(+)-(A) 5, A:ASS and from this (on introduction of a model taking only structures with truth tables as real) we can conclude that A holds as required.
  9. I can't copy and paste here, because there are many figures in the text that does not show.
  10. Sorry, I can't copy and paste here because there are many figures that does not show.
  11. The file is attached. Proof of Axioms of Propositional Logic wo name.pdf
  12. Please comment on the attached file. Please help me to prove the Logic as sound. I will include your name as co-writer. Knowledge_Organization 4 wo name.pdf
  13. What is the wave function of an electron in a Hydrogen atom?
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