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 coecients 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 dier 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.