Page:An Investigation of the Laws of Thought (1854, Boole, investigationofl00boolrich).djvu/144

Whence there will result $\scriptstyle {stq(1-r)=0,\quad str(1-q)=0}$, these being equations which are deducible from either of the primitive ones. Their interpretations are—

Similar triangles which have their corresponding angles equal have their corresponding sides proportional.

Similar triangles which have their corresponding sides proportional have their corresponding angles equal.

And these conclusions are equally deducible from either premiss singly. In this respect, according to the definitions laid down, the premises are not independent.

14. Let us, in conclusion, resume the problem discussed in illustration of the first method of this chapter, and endeavour to ascertain, by the present method, what may be concluded from the presence of the property $\scriptstyle {C}$, with reference to the properties $\scriptstyle {A}$ and $\scriptstyle {B}$.

We found on eliminating the symbols $\scriptstyle {v}$ the following equations, viz.: (1)$\scriptstyle {xy(wz+{\bar {w}}{\bar {z}})=0}$,(2)$\scriptstyle {yz(x{\bar {w}}+{\bar {x}}w)=0}$,(3)$\scriptstyle {{\bar {x}}{\bar {y}}={\bar {w}}{\bar {z}}}$. From these we are to eliminate $\scriptstyle {w}$ and determine $\scriptstyle {z}$. Now (1) and (2) already satisfy the condition $\scriptstyle {V(1-V)=0}$. The third equation gives, on bringing the terms to the first side, and squaring (4)$\scriptstyle {{\bar {x}}{\bar {y}}(1-{\bar {\nu }}{\bar {z}})+{\bar {w}}{\bar {z}}(1-{\bar {x}}{\bar {y}})=0}$^{[errata 1]}. (4) Adding (1) (2) and (4) together, we have $\scriptstyle {xy(wz+{\bar {w}}{\bar {z}})+yz(x{\bar {w}}+x)+{\bar {x}}{\bar {y}}(l-{\bar {w}}{\bar {z}})+{\bar {w}}{\bar {z}}(1-{\bar {x}}{\bar {y}})=0}$^{[errata 2]}. Eliminating $\scriptstyle {w}$, we get $\scriptstyle {(xyz+yz{\bar {x}}+{\bar {w}}{\bar {y}})\{xy{\bar {z}}+yzx+{\bar {x}}{\bar {y}}z+{\bar {z}}(1-{\bar {x}}{\bar {y}})\}=0}$^{[errata 3]}. Now, on multiplying the terms in the second factor by those in the first successively, observing that $\scriptstyle {x{\bar {x}}=0,\quad y{\bar {y}}=0,\quad z{\bar {z}}=0}$,

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1. Correction: $\scriptstyle {{\bar {x}}{\bar {y}}(1-{\bar {\nu }}{\bar {z}})+{\bar {w}}{\bar {z}}(1-{\bar {x}}{\bar {y}})=0}$ should be amended to $\scriptstyle {{\bar {x}}{\bar {y}}(1-{\bar {w}}{\bar {z}})+{\bar {w}}{\bar {z}}(1-{\bar {x}}{\bar {y}})=0}$: detail
2. Correction: $\scriptstyle {xy(wz+{\bar {w}}{\bar {z}})+yz(x{\bar {w}}+x)+{\bar {x}}{\bar {y}}(l-{\bar {w}}{\bar {z}})+{\bar {w}}{\bar {z}}(1-{\bar {x}}{\bar {y}})=0}$ should be amended to $\scriptstyle {xy(wz+{\bar {w}}{\bar {z}})+yz(x{\bar {w}}+{\bar {x}}w)+{\bar {x}}{\bar {y}}(l-{\bar {w}}{\bar {z}})+{\bar {w}}{\bar {z}}(1-{\bar {x}}{\bar {y}})=0}$: detail
3. Correction: $\scriptstyle {(xyz+yz{\bar {x}}+{\bar {w}}{\bar {y}})\{xy{\bar {z}}+yzx+{\bar {x}}{\bar {y}}z+{\bar {z}}(1-{\bar {x}}{\bar {y}})\}=0}$ should be amended to $\scriptstyle {(xyz+yz{\bar {x}}+{\bar {x}}{\bar {y}})\{xy{\bar {z}}+yzx+{\bar {x}}{\bar {y}}z+{\bar {z}}(1-{\bar {x}}{\bar {y}})\}=0}$: detail