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By Couturat L.

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If, on the other hand, we wish to eliminate x, y, z, . . , the constituents p1 , p2 , p3 . ), we put the equation in the form (A + a t + at )p1 + (B + b t + bt )p2 + (C + c t + ct )p3 + . . = 0, and the resultant will be (A + a t + at )(B + b t + bt )(C + c t + ct ) . . = 0, 47 Laws of Thought, Chap. IX, 8. 51 an equation that contains only the unknown quantity t and the constants of the problem (the coecients of f and of ϕ). From this may be derived the expression of t in terms of these constants.

K + u(a + b + c + . . + k), it will be equal to its minimum (abc . . k) when u = 0, and to its maximum (a + b + c + . . + k) when u = 1. Moreover we can verify this proposition on the primitive form of the function by giving suitable values to the variables. Thus a function can assume all values comprised between its two limits, including the limits themselves. Consequently, it is absolutely indeterminate when abc . . k = 0 and a + b + c + . . + k = 1 at the same time, or abc . . k = 0 = a b c .

Stanley Jevons, Pure Logic, 1864, p. 61. 29 other equalities which dier from the rst only in the permutation of these three letters be similarly transformed, the same result will be obtained, which proves the proposed equivalence. If we have at the same time the three inclusions: a < bc + b c, b < ac + a c, c < ab + a b. we have also the converse inclusion, an therefore the corresponding equalities a = bc + b c, b = ac + a c, c = ab + a b. For if we transform the given inclusions into equalities, we shall have abc + ab c = 0, abc + a bc = 0, abc + a b c = 0, whence, by combining them into a single equality, abc + ab c + a bc + a b c = 0.