Dear Helmut, I go back to my writings of last year and I reread the reasoning. I keepin a whole all the last two mail:
1 MAIL: Last year I read part of the book of Peano [1]. In this book Peano explains the state of art of logic in 1888. He explains in this way the rudimental concept of implication: [link to pag 9 of book <https://books.google.it/books?hl=en&lr=&id=5LJi3dxLzuwC&oi=fnd&pg=PA9&dq=peano+calcolo+geometrico&ots=4xikZ7toBC&sig=yvGzsGadG6UgBZ7pBHp9SA6wvLs&redir_esc=y#v=onepage&q=peano%20calcolo%20geometrico&f=false> ] a < b or b > a the class [proposition] defined by the condition a is part of by those defined by b, or in another way a has as a consequence b a = b if a is true and also b, and viceversa a ^ b the condition assuming that both a and b are true a U b the condition assuming that or a or b are true (a) the condition that we obtain negating a F the absurd condition T the identical condition Than the book explains the calculus of proposition and terminates with this 4 type of proposition: [link to pag 14 of book <https://books.google.it/books?hl=en&lr=&id=5LJi3dxLzuwC&oi=fnd&pg=PA14&dq=peano+calcolo+geometrico&ots=4xikZ7toBC&sig=yvGzsGadG6UgBZ7pBHp9SA6wvLs&redir_esc=y#v=onepage&q=peano%20calcolo%20geometrico&f=false> ] I) All a are b II) No a is b III) Some a is b IV) Some a is not b And he transforms the first proposition in a ^ (b) = F that is more similar at (a(b)) the cactus formula for implication Peano named these propositions in this way: The I) and II) are Universal. The III) and IV) that are negations of universal preposition, he named them particular. The I) and the III) that contain an even number of negations, he named them proposition affirmative. The II) and IV) that contains an odd number of negations, he named them negative. 2 MAIL: Dear Helmut, I'm not sure to have understood what you have said. Let's: A={n: n=4*i con i (1..infinity)} B={n: n=2*i con i (1..infinity)} I see that all a are also b. But at one moment I will see that there are some b, like for example 6,that are not a. So the not existence of a that are not b and the existence of b that are not a, drive me to conclude that A is included in B and A implies B. So if..then come after negation. It's right? NOW: so we can say that not only a->b is All A are B but also Some B is not A. We can write: [book pag 14 <https://books.google.it/books?hl=en&lr=&id=5LJi3dxLzuwC&oi=fnd&pg=PA14&dq=peano+calcolo+geometrico&ots=4xikZ7toBC&sig=yvGzsGadG6UgBZ7pBHp9SA6wvLs&redir_esc=y#v=onepage&q=peano%20calcolo%20geometrico&f=false> ] Some B is not A: ([B ^ A] =F) remember that the square brackets are separation and the brackets "()" are negation. Now we can write: (a^(b)) ^ (b^a) This is a new concept of implication: we can prove say that is included in implication concept more abstract: ([(a^(b)) ^ (b^a)] ([(a(b))])) I rewrite this in another notation. Put the sign "->" as implication: ((a^(b)) ^ (b^a)) -> (a->b) ((a->b)^(b^a))->(a->b) This is a tautology In few word: implication is: All A are B and some B are not A regards Mauro
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