talanum46 Posted June 23 Share Posted June 23 (edited) 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. Sources differ asto the axioms they choose but some of the most famous "axioms" are proved. Modus Ponens occurs in all systems. 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 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). where "<->" means: "is equivalent to" or "follows from and vice vesa". A:AD reads as follows: ((A)->-(B))->-( <-> )->-(A) []->-(B)->-( where "[]->-" is a Stopper carrying "therefore" relation. 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) where the Attractors may carry a relation symbol. 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 trowards each other and attached to two different structures. 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). In 2 "(B -> C)" is a structure. 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 (by choosing to place the Stopper on the other side of "-(x)-"). 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, truth table 3 )-(+)-(A) []-(+)-((A)-(+)-(_))-(+)-( 2, A:AtI, A:AD 4 (A) []-(+)-((A)-(+)-(_)) 3, A:ASS, A:SD, A:ASS 5 (A)-(+)-( []-(+)- )-(+)-(A) []-(+)-(_)-(+)-( 4, A:AtI, A:AD 6 (A)-(+)-( )-(+)-(A) []-(+)-(_)-(+)-( 5, T:ANDE 7 (A) [](+)-(_)-(+)-( 6, A:AOA 8 A 7, A:EED where "(_)" is the empty structure (a structure that is always false). Line 6 is because we can write line 5 as: (A)-(+)-( )-(+)-(A) []-(+)-(_)-(+)-( AND (A)-(+)-( []-(+)-(A) []-(+)-(_)-(+)-(, since they have the same meaning. 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 The primitive proposition of Principia Mathematica called "Sum." can be proven using a version of A:AA and the definition of "therefore". It needs a somewhat different version of A:AA stated as follows: A:AAE: (P)->( Q )->(P) R <> Q R Theorem 2.0.11.2: ( P -> Q ) -> ((~R OR P) -> (~R OR Q)) Proof: Line # Statement Reason 1 P -> Q Premise 2 )->-(P) []->-(Q)->-( 1, A:AtI, A:AD 3 (R)->-( )->-(P) )->-(R) []->-(Q)->-( 2, A:AAE 4 (R->P) )->-(R) []->-(Q)->-( 3, A:AL 5 (R->P) []->-(R) )->-(Q) 4, A:ASS, A:SD 6 (R->P) []->-(R)->(Q) 5, T:AL 7 (R->P) []->-((R)->(Q)) 6, See text 8 (R->P) )->-((R)->(Q)) 7, A:ASS 9 (R->P)->(R->Q) 8, T:AL 10 (~R OR P)->(~R OR Q) 9, Def. of -> where "See text" is: because the other priority assignment is refuted by the refuted axiom that follows: RA:TP: ... []->-(R)->(Q) X<> (... []->-(R))->(Q) where "X" means "not". end Edited June 23 by talanum46 minor errors Link to comment Share on other sites More sharing options...
dimreepr Posted June 23 Share Posted June 23 Well, wadda ya know, proof that a paradox can't exist, congratulations on your logical side of reason, where x = bollox... Link to comment Share on other sites More sharing options...
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