Page:An Investigation of the Laws of Thought (1854, Boole, investigationofl00boolrich).djvu/135

and in these expressions replace, for simplicity, $\scriptstyle {1-x}$ by $\scriptstyle {\bar {x}}$, $\scriptstyle {1-y}$ by $\scriptstyle {\bar {y}}$, &c., we shall have from the three last equations, (1)$\scriptstyle {xy(wz+{\bar {w}}{\bar {z}})=0}$;(2)$\scriptstyle {yz(x{\bar {w}}+{\bar {x}}w)=0}$;(3)$\scriptstyle {{\bar {x}}{\bar {y}}={\bar {v}}{\bar {z}}}$^{[errata 1]}; and from this system we must eliminate $\scriptstyle {w}$.

Multiplying the second of the above equations by $\scriptstyle {c}$, and the third by $\scriptstyle {c'}$, and adding the results to the first, we have $\scriptstyle {xy(wz+{\bar {w}}{\bar {z}})+cyz(x{\bar {w}}+{\bar {x}}w)+c'({\bar {x}}{\bar {y}}-{\bar {w}}{\bar {x}})=0}$^{[errata 2]}. When $\scriptstyle {w}$ is made equal to $\scriptstyle {1}$, and therefore $\scriptstyle {\bar {w}}$ to $\scriptstyle {0}$, the first member of the above equation becomes $\scriptstyle {xyz+c{\bar {x}}yz+c'{\bar {x}}{\bar {y}}}$. And when in the same member $\scriptstyle {w}$ is made $\scriptstyle {0}$ and $\scriptstyle {{\bar {w}}=1}$, it becomes $\scriptstyle {xy{\bar {z}}+cxyz+c'{\bar {x}}{\bar {y}}-c'{\bar {z}}}$. Hence the result of the elimination of $\scriptstyle {w}$ may be expressed in the form (4)$\scriptstyle {(xyz+c{\bar {x}}yz+c'{\bar {x}}{\bar {y}})(xy{\bar {z}}+cxyz+c'{\bar {x}}{\bar {y}}-c'{\bar {z}})=0}$; and from this equation $\scriptstyle {x}$ is to be determined.

Were we now to proceed as in former instances, we should multiply together the factors in the first member of the above equation; but it may be well to show that such a course is not at all necessary. Let us develop the first member of (4) with reference to $\scriptstyle {x}$, the symbol whose expression is sought, we find $\scriptstyle {yz(y{\bar {z}}+cyz-c'{\bar {z}})x+(cyz+c'{\bar {y}})(c'{\bar {y}}-c'{\bar {z}})(1-x)=0}$; or,$\scriptstyle {cyzx+(cyz+c'{\bar {y}})(c'{\bar {y}}-c'{\bar {z}})(1-x)=0}$; whence we find, $\scriptstyle {x={\frac {(cyz+c'{\bar {y}})(c'{\bar {y}}-c'{\bar {z}})}{(cyz+c'{\bar {y}})(c'{\bar {y}}-c'{\bar {z}})-cyz}}}$; and developing the second member with respect to $\scriptstyle {y}$ and $\scriptstyle {z}$,

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1. Correction: $\scriptstyle {{\bar {x}}{\bar {y}}={\bar {v}}{\bar {z}}}$ should be amended to $\scriptstyle {{\bar {x}}{\bar {y}}={\bar {w}}{\bar {z}}}$: detail
2. Correction: $\scriptstyle {xy(wz+{\bar {w}}{\bar {z}})+cyz(x{\bar {w}}+{\bar {x}}w)+c'({\bar {x}}{\bar {y}}-{\bar {w}}{\bar {x}})=0}$ should be amended to $\scriptstyle {xy(wz+{\bar {w}}{\bar {z}})+cyz(x{\bar {w}}+{\bar {x}}w)+c'({\bar {x}}{\bar {y}}-{\bar {w}}{\bar {z}})=0}$: detail