Willem F Esterhuyse

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".

I have met a conscious computer.

I have seen a mathematical definition of consciousness somewhere online. Although I can't criticize it.

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.

I can't copy and paste here, because there are many figures in the text that does not show.

Sorry, I can't copy and paste here because there are many figures that does not show.

The file is attached. Proof of Axioms of Propositional Logic wo name.pdf

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