A Survey of Symbolic Logic
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This method of proving the Distributive Law is taken from Huntington, "Sets of Independent Postulates for the Algebra of Logic ". The proof of the long and difficult lemma 3 is due to Peirce, who worked it out for his paper of 1880 but mislaid the sheets, and it was printed for the first time in Huntington's paper. 6 5-51 (a + b)(c + d) = (a c + b c) + (a d + b d).
[5 5] (a + 6) (c + d) = (a + 6) c + (a + 6) d = (a c + b c) + (a d + b d) .
5-52 a + bc = (a + 6)(a + c). (Correlate of 5 5) [5-51]... (a + 6) (a + c) = (a a + b a) + (a c + b c) = [(a + a 6) + a c] + &*e.
But [5-4] (a + a 6) + a c = a + a c = a. Hence Q. E. D. Further theorems which are often useful in working the algebra and which follow readily from the preceding are as follows: 6 See "Sets of Independent Postulates, etc. ", loc. C, it. , p. 300, footnote. 10 130 A Survey of Symbolic Logic 5-6 a-1 = a = 1-a.
[1-5] a-0 = 0. Hence a--l = 0. But [1-61] if a--l = 0, then a-1 = a.
5-61 acl.
[1-9] Since a-1 = a, acl. 5-62 a + = a = + a.
What to read after A Survey of Symbolic Logic?
You can find similar books in the "Read Also" column, or choose other free books by Clarence Irving Lewis to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
This method of proving the Distributive Law is taken from Huntington, "Sets of Independent Postulates for the Algebra of Logic ". The proof of the long and difficult lemma 3 is due to Peirce, who worked it out for his paper of 1880 but mislaid the sheets, and it was printed for the first time in Huntington's paper. 6 5-51 (a + b)(c + d) = (a c + b c) + (a d + b d).
[5 5] (a + 6) (c + d) = (a + 6) c + (a + 6) d = (a c + b c) + (a d + b d) .
5-52 a + bc = (a + 6)(a + c). (Correlate of 5 5) [5-51]... (a + 6) (a + c) = (a a + b a) + (a c + b c) = [(a + a 6) + a c] + &*e.
But [5-4] (a + a 6) + a c = a + a c = a. Hence Q. E. D. Further theorems which are often useful in working the algebra and which follow readily from the preceding are as follows: 6 See "Sets of Independent Postulates, etc. ", loc. C, it. , p. 300, footnote. 10 130 A Survey of Symbolic Logic 5-6 a-1 = a = 1-a.
[1-5] a-0 = 0. Hence a--l = 0. But [1-61] if a--l = 0, then a-1 = a.
5-61 acl.
[1-9] Since a-1 = a, acl. 5-62 a + = a = + a.
What to read after A Survey of Symbolic Logic?
You can find similar books in the "Read Also" column, or choose other free books by Clarence Irving Lewis to read onlineMoreLess
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